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?
$begingroup$
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 $x$?
limits tangent-line slope
$endgroup$
add a comment |
$begingroup$
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 $x$?
limits tangent-line slope
$endgroup$
$begingroup$
"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
add a comment |
$begingroup$
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 $x$?
limits tangent-line slope
$endgroup$
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
limits tangent-line slope
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
add a comment |
$begingroup$
"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
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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.
This is not achieved with a different slope.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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
$$f(x) = f'(x_0)(x-x_0) + f(x_0).$$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$endgroup$
add a comment |
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4 Answers
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active
oldest
votes
4 Answers
4
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
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.
This is not achieved with a different slope.
$endgroup$
add a comment |
$begingroup$
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.
This is not achieved with a different slope.
$endgroup$
add a comment |
$begingroup$
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.
This is not achieved with a different slope.
$endgroup$
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.
This is not achieved with a different slope.
edited 1 hour ago
answered 1 hour ago
Yves DaoustYves Daoust
133k676232
133k676232
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 1 hour ago
AllawonderAllawonder
2,304617
2,304617
add a comment |
add a comment |
$begingroup$
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
$$f(x) = f'(x_0)(x-x_0) + f(x_0).$$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$f(x) = f'(x_0)(x-x_0) + f(x_0).$$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$f(x) = f'(x_0)(x-x_0) + f(x_0).$$
$endgroup$
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
$$f(x) = f'(x_0)(x-x_0) + f(x_0).$$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
edited 51 mins ago
answered 1 hour ago
yousef magablehyousef magableh
429
429
add a comment |
add a comment |
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$begingroup$
"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