Why limits give us the exact value of the slope of the tangent line? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Why doesn't derivative difference quotient violate the epsilon-delta definition of a limit?The slope of the tangent lineGive equation of g(t) = |t-2| / t and sketch a graph of slope of the tangent line as t varies from 0 to 4.Why find the slope other than finding min , max?How to write an equation of a line bisecting an angle in terms of the slope of the bisector.Is the definition of a tangent line flawed?How tangent line gives us a slope at one point?Definition of a Tangent line and DerivativesFind the tangent line for the following: $(operatornamearcsec x)^2$ at $x = 2$Why Is The Derivative At A Point Drawn As A Tangent Line?How can I find the slope of a line tangent to a small circle on a sphere?

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Why limits give us the exact value of the slope of the tangent line?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Why doesn't derivative difference quotient violate the epsilon-delta definition of a limit?The slope of the tangent lineGive equation of g(t) = |t-2| / t and sketch a graph of slope of the tangent line as t varies from 0 to 4.Why find the slope other than finding min , max?How to write an equation of a line bisecting an angle in terms of the slope of the bisector.Is the definition of a tangent line flawed?How tangent line gives us a slope at one point?Definition of a Tangent line and DerivativesFind the tangent line for the following: $(operatornamearcsec x)^2$ at $x = 2$Why Is The Derivative At A Point Drawn As A Tangent Line?How can I find the slope of a line tangent to a small circle on a sphere?










4












$begingroup$


Limits tell us how functions behave at $xto a$, not how they behave at $x = a$. However, in limits we plug $x = a$ as an approximation of $xto a$, so: why the limits give us the exact value of slope of the tangent line, despite being just an approximation of what happend around $x$?










share|cite|improve this question











$endgroup$











  • $begingroup$
    "But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
    $endgroup$
    – Dave L. Renfro
    2 hours ago











  • $begingroup$
    First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
    $endgroup$
    – Allawonder
    1 hour ago















4












$begingroup$


Limits tell us how functions behave at $xto a$, not how they behave at $x = a$. However, in limits we plug $x = a$ as an approximation of $xto a$, so: why the limits give us the exact value of slope of the tangent line, despite being just an approximation of what happend around $x$?










share|cite|improve this question











$endgroup$











  • $begingroup$
    "But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
    $endgroup$
    – Dave L. Renfro
    2 hours ago











  • $begingroup$
    First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
    $endgroup$
    – Allawonder
    1 hour ago













4












4








4





$begingroup$


Limits tell us how functions behave at $xto a$, not how they behave at $x = a$. However, in limits we plug $x = a$ as an approximation of $xto a$, so: why the limits give us the exact value of slope of the tangent line, despite being just an approximation of what happend around $x$?










share|cite|improve this question











$endgroup$




Limits tell us how functions behave at $xto a$, not how they behave at $x = a$. However, in limits we plug $x = a$ as an approximation of $xto a$, so: why the limits give us the exact value of slope of the tangent line, despite being just an approximation of what happend around $x$?







limits tangent-line slope






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









BowPark

567720




567720










asked 3 hours ago









Mohammad AlshareefMohammad Alshareef

264




264











  • $begingroup$
    "But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
    $endgroup$
    – Dave L. Renfro
    2 hours ago











  • $begingroup$
    First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
    $endgroup$
    – Allawonder
    1 hour ago
















  • $begingroup$
    "But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
    $endgroup$
    – Dave L. Renfro
    2 hours ago











  • $begingroup$
    First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
    $endgroup$
    – Allawonder
    1 hour ago















$begingroup$
"But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
$endgroup$
– Dave L. Renfro
2 hours ago





$begingroup$
"But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
$endgroup$
– Dave L. Renfro
2 hours ago













$begingroup$
First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
$endgroup$
– Allawonder
1 hour ago




$begingroup$
First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
$endgroup$
– Allawonder
1 hour ago










4 Answers
4






active

oldest

votes


















1












$begingroup$

The idea is that the approximation gets better and better as $x$ gets closer to $a$, and, so to say, at $x=a$ it is no more approximative.



Below, a function in blue, the tangent in green and the difference in magenta. As you see, the difference decreases very quickly close to the tangency point.



enter image description here



This is not achieved with a different slope.



enter image description here






share|cite|improve this answer











$endgroup$




















    1












    $begingroup$

    A line by line correction of OP that should answer the confusions therein:



    (1) Limits tell us how a function behaves nearby $x=a.$ The function doesn't even have to be defined at $a.$ If it is also defined there, then the limit tells us in addition what's happening at $a,$ contrary to your first sentence. But the important thing to note is that limits tell us how a function is behaving in the neighborhood of some point in its domain.



    (2) From above, it follows that $f(a)$ need not exist at all in order for $lim_xto af(x)$ to exist. Indeed, even if $f(a)$ exists, it need not coincide with the limiting value. In essence, limiting values have nothing essentially to do with values. But in cases where $lim_xto af(x)$ exists and coincides with $f(a),$ we may use this fact to evaluate the limiting value, but it is no way an approximation. There's nothing about approximations here. The conditions are all equalities. Again, the salient point here is that we can't always evaluate limits, when they exist, by a substitution as you think. This only applies to functions continuous at the point in question.



    (3) You have yet to understand the concept of limit properly. It is not an approximation to anything, but rather that towards which we approximate; thus by definition it will sometimes transcend every approximant. Rather, limits are a way to extend usual concepts into wider domains of applicability in such a way that the new objects properly generalise the old -- that is, they possess all important properties of the old objects, and contain the latter as a subclass. In particular here, bringing in limiting operations helps us to assign a definite meaning to some sequences of approximations that distinguish themselves from others. Note that whereas some terms of these types of sequence approximate the limiting value, this value itself is decidedly not an approximation, but a uniquely defined value, fixed by this sequence (and an assumption about nonempty subsets of real numbers that genuinely and smoothly extends the concept of maximum or minimum values of nonempty subsets of numbers). This well-defined limiting value, whenever it exists, has been shown many times over to legitimately (in a sense that can be made precise) extend the domain of application of the concept under study. Here, it follows that whenever a sequence of approximating slopes converges, we can choose to call (by a justified extension) the limit a slope too, indeed the slope, defined by that sequence. That's simply what it means to work with real numbers. If you've not understood this property (technically called completeness) then you've not begun to understand the real numbers.






    share|cite|improve this answer









    $endgroup$




















      0












      $begingroup$

      Well, if you have two points $(x_1,y_1)$ and $(x_2,y_2)$ in the plane, then the slope of the unique line through these points is
      $$ m = fracy_2-y_1x_2-x_1.$$
      If you consider a (differentiable) function $f:Bbb RrightarrowBbb R$, then the slope of the line between $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is
      $$ m = fracf(x_2)-f(x_1)x_2-x_1.$$
      If you take the points $x_0$ and $x$ at the x-axis, then consider the quotient
      $$fracf(x)-f(x_0)x-x_0$$
      when $x$ moves to $x_0$: $xrightarrow x_0$.
      Then this quotient becomes (in the limit) the slope of the tangent line of $f$ at the point $x_0$. This is particularly easy to understand if you draw a picture.



      Denote the slope in the limit as $f'(x_0)$. Then the above quotient gives
      $$f'(x_0) = fracf(x)-f(x_0)x-x_0$$
      and so the equation of the tangent line
      $$f(x) = f'(x_0)(x-x_0) + f(x_0).$$






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
        $endgroup$
        – Mohammad Alshareef
        3 hours ago







      • 1




        $begingroup$
        @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
        $endgroup$
        – Dave L. Renfro
        2 hours ago










      • $begingroup$
        @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
        $endgroup$
        – Mohammad Alshareef
        2 hours ago











      • $begingroup$
        Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
        $endgroup$
        – Dave L. Renfro
        2 hours ago











      • $begingroup$
        Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
        $endgroup$
        – Mohammad Alshareef
        2 hours ago



















      0












      $begingroup$

      I think the problem with your thinking is that, you are assuming that there is something that is predefined by "god" which is the slope at a point that has exact value, and we are trying to reach it by the processes of taking a limit, and we can never reach that value. But that is not the case. What we do, is that we define the notion of the slope at a point , by the processes of taking a limit. And what ever the exact value that comes out of the limit, is what we call the slope at that point.



      I guess that what is bothering you, is the notion of a limit that you learn at calculus , and i felt the same when i first learn it, because at some stage of the problem we say that x approaches some value a , then at another stage we substitute a for x. But to understand what is actually going on, you have to understand the epsilon-delta definition of a limit.
      see:https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit






      share|cite|improve this answer











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        4 Answers
        4






        active

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        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        The idea is that the approximation gets better and better as $x$ gets closer to $a$, and, so to say, at $x=a$ it is no more approximative.



        Below, a function in blue, the tangent in green and the difference in magenta. As you see, the difference decreases very quickly close to the tangency point.



        enter image description here



        This is not achieved with a different slope.



        enter image description here






        share|cite|improve this answer











        $endgroup$

















          1












          $begingroup$

          The idea is that the approximation gets better and better as $x$ gets closer to $a$, and, so to say, at $x=a$ it is no more approximative.



          Below, a function in blue, the tangent in green and the difference in magenta. As you see, the difference decreases very quickly close to the tangency point.



          enter image description here



          This is not achieved with a different slope.



          enter image description here






          share|cite|improve this answer











          $endgroup$















            1












            1








            1





            $begingroup$

            The idea is that the approximation gets better and better as $x$ gets closer to $a$, and, so to say, at $x=a$ it is no more approximative.



            Below, a function in blue, the tangent in green and the difference in magenta. As you see, the difference decreases very quickly close to the tangency point.



            enter image description here



            This is not achieved with a different slope.



            enter image description here






            share|cite|improve this answer











            $endgroup$



            The idea is that the approximation gets better and better as $x$ gets closer to $a$, and, so to say, at $x=a$ it is no more approximative.



            Below, a function in blue, the tangent in green and the difference in magenta. As you see, the difference decreases very quickly close to the tangency point.



            enter image description here



            This is not achieved with a different slope.



            enter image description here







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago

























            answered 1 hour ago









            Yves DaoustYves Daoust

            133k676232




            133k676232





















                1












                $begingroup$

                A line by line correction of OP that should answer the confusions therein:



                (1) Limits tell us how a function behaves nearby $x=a.$ The function doesn't even have to be defined at $a.$ If it is also defined there, then the limit tells us in addition what's happening at $a,$ contrary to your first sentence. But the important thing to note is that limits tell us how a function is behaving in the neighborhood of some point in its domain.



                (2) From above, it follows that $f(a)$ need not exist at all in order for $lim_xto af(x)$ to exist. Indeed, even if $f(a)$ exists, it need not coincide with the limiting value. In essence, limiting values have nothing essentially to do with values. But in cases where $lim_xto af(x)$ exists and coincides with $f(a),$ we may use this fact to evaluate the limiting value, but it is no way an approximation. There's nothing about approximations here. The conditions are all equalities. Again, the salient point here is that we can't always evaluate limits, when they exist, by a substitution as you think. This only applies to functions continuous at the point in question.



                (3) You have yet to understand the concept of limit properly. It is not an approximation to anything, but rather that towards which we approximate; thus by definition it will sometimes transcend every approximant. Rather, limits are a way to extend usual concepts into wider domains of applicability in such a way that the new objects properly generalise the old -- that is, they possess all important properties of the old objects, and contain the latter as a subclass. In particular here, bringing in limiting operations helps us to assign a definite meaning to some sequences of approximations that distinguish themselves from others. Note that whereas some terms of these types of sequence approximate the limiting value, this value itself is decidedly not an approximation, but a uniquely defined value, fixed by this sequence (and an assumption about nonempty subsets of real numbers that genuinely and smoothly extends the concept of maximum or minimum values of nonempty subsets of numbers). This well-defined limiting value, whenever it exists, has been shown many times over to legitimately (in a sense that can be made precise) extend the domain of application of the concept under study. Here, it follows that whenever a sequence of approximating slopes converges, we can choose to call (by a justified extension) the limit a slope too, indeed the slope, defined by that sequence. That's simply what it means to work with real numbers. If you've not understood this property (technically called completeness) then you've not begun to understand the real numbers.






                share|cite|improve this answer









                $endgroup$

















                  1












                  $begingroup$

                  A line by line correction of OP that should answer the confusions therein:



                  (1) Limits tell us how a function behaves nearby $x=a.$ The function doesn't even have to be defined at $a.$ If it is also defined there, then the limit tells us in addition what's happening at $a,$ contrary to your first sentence. But the important thing to note is that limits tell us how a function is behaving in the neighborhood of some point in its domain.



                  (2) From above, it follows that $f(a)$ need not exist at all in order for $lim_xto af(x)$ to exist. Indeed, even if $f(a)$ exists, it need not coincide with the limiting value. In essence, limiting values have nothing essentially to do with values. But in cases where $lim_xto af(x)$ exists and coincides with $f(a),$ we may use this fact to evaluate the limiting value, but it is no way an approximation. There's nothing about approximations here. The conditions are all equalities. Again, the salient point here is that we can't always evaluate limits, when they exist, by a substitution as you think. This only applies to functions continuous at the point in question.



                  (3) You have yet to understand the concept of limit properly. It is not an approximation to anything, but rather that towards which we approximate; thus by definition it will sometimes transcend every approximant. Rather, limits are a way to extend usual concepts into wider domains of applicability in such a way that the new objects properly generalise the old -- that is, they possess all important properties of the old objects, and contain the latter as a subclass. In particular here, bringing in limiting operations helps us to assign a definite meaning to some sequences of approximations that distinguish themselves from others. Note that whereas some terms of these types of sequence approximate the limiting value, this value itself is decidedly not an approximation, but a uniquely defined value, fixed by this sequence (and an assumption about nonempty subsets of real numbers that genuinely and smoothly extends the concept of maximum or minimum values of nonempty subsets of numbers). This well-defined limiting value, whenever it exists, has been shown many times over to legitimately (in a sense that can be made precise) extend the domain of application of the concept under study. Here, it follows that whenever a sequence of approximating slopes converges, we can choose to call (by a justified extension) the limit a slope too, indeed the slope, defined by that sequence. That's simply what it means to work with real numbers. If you've not understood this property (technically called completeness) then you've not begun to understand the real numbers.






                  share|cite|improve this answer









                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    A line by line correction of OP that should answer the confusions therein:



                    (1) Limits tell us how a function behaves nearby $x=a.$ The function doesn't even have to be defined at $a.$ If it is also defined there, then the limit tells us in addition what's happening at $a,$ contrary to your first sentence. But the important thing to note is that limits tell us how a function is behaving in the neighborhood of some point in its domain.



                    (2) From above, it follows that $f(a)$ need not exist at all in order for $lim_xto af(x)$ to exist. Indeed, even if $f(a)$ exists, it need not coincide with the limiting value. In essence, limiting values have nothing essentially to do with values. But in cases where $lim_xto af(x)$ exists and coincides with $f(a),$ we may use this fact to evaluate the limiting value, but it is no way an approximation. There's nothing about approximations here. The conditions are all equalities. Again, the salient point here is that we can't always evaluate limits, when they exist, by a substitution as you think. This only applies to functions continuous at the point in question.



                    (3) You have yet to understand the concept of limit properly. It is not an approximation to anything, but rather that towards which we approximate; thus by definition it will sometimes transcend every approximant. Rather, limits are a way to extend usual concepts into wider domains of applicability in such a way that the new objects properly generalise the old -- that is, they possess all important properties of the old objects, and contain the latter as a subclass. In particular here, bringing in limiting operations helps us to assign a definite meaning to some sequences of approximations that distinguish themselves from others. Note that whereas some terms of these types of sequence approximate the limiting value, this value itself is decidedly not an approximation, but a uniquely defined value, fixed by this sequence (and an assumption about nonempty subsets of real numbers that genuinely and smoothly extends the concept of maximum or minimum values of nonempty subsets of numbers). This well-defined limiting value, whenever it exists, has been shown many times over to legitimately (in a sense that can be made precise) extend the domain of application of the concept under study. Here, it follows that whenever a sequence of approximating slopes converges, we can choose to call (by a justified extension) the limit a slope too, indeed the slope, defined by that sequence. That's simply what it means to work with real numbers. If you've not understood this property (technically called completeness) then you've not begun to understand the real numbers.






                    share|cite|improve this answer









                    $endgroup$



                    A line by line correction of OP that should answer the confusions therein:



                    (1) Limits tell us how a function behaves nearby $x=a.$ The function doesn't even have to be defined at $a.$ If it is also defined there, then the limit tells us in addition what's happening at $a,$ contrary to your first sentence. But the important thing to note is that limits tell us how a function is behaving in the neighborhood of some point in its domain.



                    (2) From above, it follows that $f(a)$ need not exist at all in order for $lim_xto af(x)$ to exist. Indeed, even if $f(a)$ exists, it need not coincide with the limiting value. In essence, limiting values have nothing essentially to do with values. But in cases where $lim_xto af(x)$ exists and coincides with $f(a),$ we may use this fact to evaluate the limiting value, but it is no way an approximation. There's nothing about approximations here. The conditions are all equalities. Again, the salient point here is that we can't always evaluate limits, when they exist, by a substitution as you think. This only applies to functions continuous at the point in question.



                    (3) You have yet to understand the concept of limit properly. It is not an approximation to anything, but rather that towards which we approximate; thus by definition it will sometimes transcend every approximant. Rather, limits are a way to extend usual concepts into wider domains of applicability in such a way that the new objects properly generalise the old -- that is, they possess all important properties of the old objects, and contain the latter as a subclass. In particular here, bringing in limiting operations helps us to assign a definite meaning to some sequences of approximations that distinguish themselves from others. Note that whereas some terms of these types of sequence approximate the limiting value, this value itself is decidedly not an approximation, but a uniquely defined value, fixed by this sequence (and an assumption about nonempty subsets of real numbers that genuinely and smoothly extends the concept of maximum or minimum values of nonempty subsets of numbers). This well-defined limiting value, whenever it exists, has been shown many times over to legitimately (in a sense that can be made precise) extend the domain of application of the concept under study. Here, it follows that whenever a sequence of approximating slopes converges, we can choose to call (by a justified extension) the limit a slope too, indeed the slope, defined by that sequence. That's simply what it means to work with real numbers. If you've not understood this property (technically called completeness) then you've not begun to understand the real numbers.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    AllawonderAllawonder

                    2,304617




                    2,304617





















                        0












                        $begingroup$

                        Well, if you have two points $(x_1,y_1)$ and $(x_2,y_2)$ in the plane, then the slope of the unique line through these points is
                        $$ m = fracy_2-y_1x_2-x_1.$$
                        If you consider a (differentiable) function $f:Bbb RrightarrowBbb R$, then the slope of the line between $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is
                        $$ m = fracf(x_2)-f(x_1)x_2-x_1.$$
                        If you take the points $x_0$ and $x$ at the x-axis, then consider the quotient
                        $$fracf(x)-f(x_0)x-x_0$$
                        when $x$ moves to $x_0$: $xrightarrow x_0$.
                        Then this quotient becomes (in the limit) the slope of the tangent line of $f$ at the point $x_0$. This is particularly easy to understand if you draw a picture.



                        Denote the slope in the limit as $f'(x_0)$. Then the above quotient gives
                        $$f'(x_0) = fracf(x)-f(x_0)x-x_0$$
                        and so the equation of the tangent line
                        $$f(x) = f'(x_0)(x-x_0) + f(x_0).$$






                        share|cite|improve this answer











                        $endgroup$












                        • $begingroup$
                          Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                          $endgroup$
                          – Mohammad Alshareef
                          3 hours ago







                        • 1




                          $begingroup$
                          @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                          $endgroup$
                          – Dave L. Renfro
                          2 hours ago










                        • $begingroup$
                          @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                          $endgroup$
                          – Mohammad Alshareef
                          2 hours ago











                        • $begingroup$
                          Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                          $endgroup$
                          – Dave L. Renfro
                          2 hours ago











                        • $begingroup$
                          Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                          $endgroup$
                          – Mohammad Alshareef
                          2 hours ago
















                        0












                        $begingroup$

                        Well, if you have two points $(x_1,y_1)$ and $(x_2,y_2)$ in the plane, then the slope of the unique line through these points is
                        $$ m = fracy_2-y_1x_2-x_1.$$
                        If you consider a (differentiable) function $f:Bbb RrightarrowBbb R$, then the slope of the line between $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is
                        $$ m = fracf(x_2)-f(x_1)x_2-x_1.$$
                        If you take the points $x_0$ and $x$ at the x-axis, then consider the quotient
                        $$fracf(x)-f(x_0)x-x_0$$
                        when $x$ moves to $x_0$: $xrightarrow x_0$.
                        Then this quotient becomes (in the limit) the slope of the tangent line of $f$ at the point $x_0$. This is particularly easy to understand if you draw a picture.



                        Denote the slope in the limit as $f'(x_0)$. Then the above quotient gives
                        $$f'(x_0) = fracf(x)-f(x_0)x-x_0$$
                        and so the equation of the tangent line
                        $$f(x) = f'(x_0)(x-x_0) + f(x_0).$$






                        share|cite|improve this answer











                        $endgroup$












                        • $begingroup$
                          Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                          $endgroup$
                          – Mohammad Alshareef
                          3 hours ago







                        • 1




                          $begingroup$
                          @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                          $endgroup$
                          – Dave L. Renfro
                          2 hours ago










                        • $begingroup$
                          @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                          $endgroup$
                          – Mohammad Alshareef
                          2 hours ago











                        • $begingroup$
                          Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                          $endgroup$
                          – Dave L. Renfro
                          2 hours ago











                        • $begingroup$
                          Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                          $endgroup$
                          – Mohammad Alshareef
                          2 hours ago














                        0












                        0








                        0





                        $begingroup$

                        Well, if you have two points $(x_1,y_1)$ and $(x_2,y_2)$ in the plane, then the slope of the unique line through these points is
                        $$ m = fracy_2-y_1x_2-x_1.$$
                        If you consider a (differentiable) function $f:Bbb RrightarrowBbb R$, then the slope of the line between $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is
                        $$ m = fracf(x_2)-f(x_1)x_2-x_1.$$
                        If you take the points $x_0$ and $x$ at the x-axis, then consider the quotient
                        $$fracf(x)-f(x_0)x-x_0$$
                        when $x$ moves to $x_0$: $xrightarrow x_0$.
                        Then this quotient becomes (in the limit) the slope of the tangent line of $f$ at the point $x_0$. This is particularly easy to understand if you draw a picture.



                        Denote the slope in the limit as $f'(x_0)$. Then the above quotient gives
                        $$f'(x_0) = fracf(x)-f(x_0)x-x_0$$
                        and so the equation of the tangent line
                        $$f(x) = f'(x_0)(x-x_0) + f(x_0).$$






                        share|cite|improve this answer











                        $endgroup$



                        Well, if you have two points $(x_1,y_1)$ and $(x_2,y_2)$ in the plane, then the slope of the unique line through these points is
                        $$ m = fracy_2-y_1x_2-x_1.$$
                        If you consider a (differentiable) function $f:Bbb RrightarrowBbb R$, then the slope of the line between $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is
                        $$ m = fracf(x_2)-f(x_1)x_2-x_1.$$
                        If you take the points $x_0$ and $x$ at the x-axis, then consider the quotient
                        $$fracf(x)-f(x_0)x-x_0$$
                        when $x$ moves to $x_0$: $xrightarrow x_0$.
                        Then this quotient becomes (in the limit) the slope of the tangent line of $f$ at the point $x_0$. This is particularly easy to understand if you draw a picture.



                        Denote the slope in the limit as $f'(x_0)$. Then the above quotient gives
                        $$f'(x_0) = fracf(x)-f(x_0)x-x_0$$
                        and so the equation of the tangent line
                        $$f(x) = f'(x_0)(x-x_0) + f(x_0).$$







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited 2 hours ago

























                        answered 3 hours ago









                        WuestenfuxWuestenfux

                        5,5561513




                        5,5561513











                        • $begingroup$
                          Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                          $endgroup$
                          – Mohammad Alshareef
                          3 hours ago







                        • 1




                          $begingroup$
                          @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                          $endgroup$
                          – Dave L. Renfro
                          2 hours ago










                        • $begingroup$
                          @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                          $endgroup$
                          – Mohammad Alshareef
                          2 hours ago











                        • $begingroup$
                          Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                          $endgroup$
                          – Dave L. Renfro
                          2 hours ago











                        • $begingroup$
                          Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                          $endgroup$
                          – Mohammad Alshareef
                          2 hours ago

















                        • $begingroup$
                          Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                          $endgroup$
                          – Mohammad Alshareef
                          3 hours ago







                        • 1




                          $begingroup$
                          @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                          $endgroup$
                          – Dave L. Renfro
                          2 hours ago










                        • $begingroup$
                          @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                          $endgroup$
                          – Mohammad Alshareef
                          2 hours ago











                        • $begingroup$
                          Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                          $endgroup$
                          – Dave L. Renfro
                          2 hours ago











                        • $begingroup$
                          Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                          $endgroup$
                          – Mohammad Alshareef
                          2 hours ago
















                        $begingroup$
                        Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                        $endgroup$
                        – Mohammad Alshareef
                        3 hours ago





                        $begingroup$
                        Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                        $endgroup$
                        – Mohammad Alshareef
                        3 hours ago





                        1




                        1




                        $begingroup$
                        @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                        $endgroup$
                        – Dave L. Renfro
                        2 hours ago




                        $begingroup$
                        @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                        $endgroup$
                        – Dave L. Renfro
                        2 hours ago












                        $begingroup$
                        @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                        $endgroup$
                        – Mohammad Alshareef
                        2 hours ago





                        $begingroup$
                        @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                        $endgroup$
                        – Mohammad Alshareef
                        2 hours ago













                        $begingroup$
                        Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                        $endgroup$
                        – Dave L. Renfro
                        2 hours ago





                        $begingroup$
                        Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                        $endgroup$
                        – Dave L. Renfro
                        2 hours ago













                        $begingroup$
                        Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                        $endgroup$
                        – Mohammad Alshareef
                        2 hours ago





                        $begingroup$
                        Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                        $endgroup$
                        – Mohammad Alshareef
                        2 hours ago












                        0












                        $begingroup$

                        I think the problem with your thinking is that, you are assuming that there is something that is predefined by "god" which is the slope at a point that has exact value, and we are trying to reach it by the processes of taking a limit, and we can never reach that value. But that is not the case. What we do, is that we define the notion of the slope at a point , by the processes of taking a limit. And what ever the exact value that comes out of the limit, is what we call the slope at that point.



                        I guess that what is bothering you, is the notion of a limit that you learn at calculus , and i felt the same when i first learn it, because at some stage of the problem we say that x approaches some value a , then at another stage we substitute a for x. But to understand what is actually going on, you have to understand the epsilon-delta definition of a limit.
                        see:https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit






                        share|cite|improve this answer











                        $endgroup$

















                          0












                          $begingroup$

                          I think the problem with your thinking is that, you are assuming that there is something that is predefined by "god" which is the slope at a point that has exact value, and we are trying to reach it by the processes of taking a limit, and we can never reach that value. But that is not the case. What we do, is that we define the notion of the slope at a point , by the processes of taking a limit. And what ever the exact value that comes out of the limit, is what we call the slope at that point.



                          I guess that what is bothering you, is the notion of a limit that you learn at calculus , and i felt the same when i first learn it, because at some stage of the problem we say that x approaches some value a , then at another stage we substitute a for x. But to understand what is actually going on, you have to understand the epsilon-delta definition of a limit.
                          see:https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit






                          share|cite|improve this answer











                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            I think the problem with your thinking is that, you are assuming that there is something that is predefined by "god" which is the slope at a point that has exact value, and we are trying to reach it by the processes of taking a limit, and we can never reach that value. But that is not the case. What we do, is that we define the notion of the slope at a point , by the processes of taking a limit. And what ever the exact value that comes out of the limit, is what we call the slope at that point.



                            I guess that what is bothering you, is the notion of a limit that you learn at calculus , and i felt the same when i first learn it, because at some stage of the problem we say that x approaches some value a , then at another stage we substitute a for x. But to understand what is actually going on, you have to understand the epsilon-delta definition of a limit.
                            see:https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit






                            share|cite|improve this answer











                            $endgroup$



                            I think the problem with your thinking is that, you are assuming that there is something that is predefined by "god" which is the slope at a point that has exact value, and we are trying to reach it by the processes of taking a limit, and we can never reach that value. But that is not the case. What we do, is that we define the notion of the slope at a point , by the processes of taking a limit. And what ever the exact value that comes out of the limit, is what we call the slope at that point.



                            I guess that what is bothering you, is the notion of a limit that you learn at calculus , and i felt the same when i first learn it, because at some stage of the problem we say that x approaches some value a , then at another stage we substitute a for x. But to understand what is actually going on, you have to understand the epsilon-delta definition of a limit.
                            see:https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 51 mins ago

























                            answered 1 hour ago









                            yousef magablehyousef magableh

                            429




                            429



























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