Is this another way of expressing the side limit? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Use the $varepsilon$-$delta$ definition of a limit to prove this.How do I solve this indeterminate limit without the L'hospital rule?Expressing a limit in different way?Proving a limit using another limitHow can I prove $lim limits_n to inftyleft(1+frac1a_nright)^a_n=e$ without involving function limit?why is $limlimits_ktoinftyfrac1+kk^k=0?$limit $ lim limits_n to infty left(fracz^1/sqrt n + z^-1/sqrt n2right)^n $How can I prove that $limlimits_x to x_0 left|fracf(x)g(x)right| = infty$ when working with punctured neighborhoods?explain why the following definition is not equivalent to the definition of the limit of function fAnother way to compute $limlimits_xto+inftyx^2logbigg(dfracx^2+1x^2+3bigg)$
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Is this another way of expressing the side limit?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Use the $varepsilon$-$delta$ definition of a limit to prove this.How do I solve this indeterminate limit without the L'hospital rule?Expressing a limit in different way?Proving a limit using another limitHow can I prove $lim limits_n to inftyleft(1+frac1a_nright)^a_n=e$ without involving function limit?why is $limlimits_ktoinftyfrac1+kk^k=0?$limit $ lim limits_n to infty left(fracz^1/sqrt n + z^-1/sqrt n2right)^n $How can I prove that $limlimits_x to x_0 left|fracf(x)g(x)right| = infty$ when working with punctured neighborhoods?explain why the following definition is not equivalent to the definition of the limit of function fAnother way to compute $limlimits_xto+inftyx^2logbigg(dfracx^2+1x^2+3bigg)$
$begingroup$
Let $f:mathbbR to mathbbR$. Can we say that $limlimits_n to inftyf(x_0+frac1n)=l$ is another way of expressing the right-sided limit at $x_0$?
I tried to use the definition of the limit,but I am stuck.Intuitively, it seems true, but I don't know how to prove it.
real-analysis limits definition
$endgroup$
add a comment |
$begingroup$
Let $f:mathbbR to mathbbR$. Can we say that $limlimits_n to inftyf(x_0+frac1n)=l$ is another way of expressing the right-sided limit at $x_0$?
I tried to use the definition of the limit,but I am stuck.Intuitively, it seems true, but I don't know how to prove it.
real-analysis limits definition
$endgroup$
add a comment |
$begingroup$
Let $f:mathbbR to mathbbR$. Can we say that $limlimits_n to inftyf(x_0+frac1n)=l$ is another way of expressing the right-sided limit at $x_0$?
I tried to use the definition of the limit,but I am stuck.Intuitively, it seems true, but I don't know how to prove it.
real-analysis limits definition
$endgroup$
Let $f:mathbbR to mathbbR$. Can we say that $limlimits_n to inftyf(x_0+frac1n)=l$ is another way of expressing the right-sided limit at $x_0$?
I tried to use the definition of the limit,but I am stuck.Intuitively, it seems true, but I don't know how to prove it.
real-analysis limits definition
real-analysis limits definition
edited 48 mins ago
GNUSupporter 8964民主女神 地下教會
14.1k82651
14.1k82651
asked 49 mins ago
Math GuyMath Guy
1407
1407
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2 Answers
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$begingroup$
No, it is not true. Take, for instance,$$f(x)=begincasessinleft(fracpi xright)&text if xneq0\0&text if x=0.endcases$$Then the limit $lim_xto0^+f(x)$ doesn't exist. However, $lim_ntoinftyfleft(frac1nright)=0$.
$endgroup$
1
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
39 mins ago
1
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
35 mins ago
add a comment |
$begingroup$
No. Right hand limit of $f$ at $x_0$ is $l$ if $f(x_0+r_n) to l$ for every sequence $(r_n)$ decreasing to $0$. A particular sequence will not do. For example, take $f(x)=0$ for $x$ rational and $1$ for $x$ irrational. Take $x_0=0$.
$endgroup$
add a comment |
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$begingroup$
No, it is not true. Take, for instance,$$f(x)=begincasessinleft(fracpi xright)&text if xneq0\0&text if x=0.endcases$$Then the limit $lim_xto0^+f(x)$ doesn't exist. However, $lim_ntoinftyfleft(frac1nright)=0$.
$endgroup$
1
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
39 mins ago
1
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
35 mins ago
add a comment |
$begingroup$
No, it is not true. Take, for instance,$$f(x)=begincasessinleft(fracpi xright)&text if xneq0\0&text if x=0.endcases$$Then the limit $lim_xto0^+f(x)$ doesn't exist. However, $lim_ntoinftyfleft(frac1nright)=0$.
$endgroup$
1
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
39 mins ago
1
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
35 mins ago
add a comment |
$begingroup$
No, it is not true. Take, for instance,$$f(x)=begincasessinleft(fracpi xright)&text if xneq0\0&text if x=0.endcases$$Then the limit $lim_xto0^+f(x)$ doesn't exist. However, $lim_ntoinftyfleft(frac1nright)=0$.
$endgroup$
No, it is not true. Take, for instance,$$f(x)=begincasessinleft(fracpi xright)&text if xneq0\0&text if x=0.endcases$$Then the limit $lim_xto0^+f(x)$ doesn't exist. However, $lim_ntoinftyfleft(frac1nright)=0$.
answered 46 mins ago
José Carlos SantosJosé Carlos Santos
176k24135244
176k24135244
1
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
39 mins ago
1
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
35 mins ago
add a comment |
1
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
39 mins ago
1
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
35 mins ago
1
1
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
39 mins ago
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
39 mins ago
1
1
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
35 mins ago
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
35 mins ago
add a comment |
$begingroup$
No. Right hand limit of $f$ at $x_0$ is $l$ if $f(x_0+r_n) to l$ for every sequence $(r_n)$ decreasing to $0$. A particular sequence will not do. For example, take $f(x)=0$ for $x$ rational and $1$ for $x$ irrational. Take $x_0=0$.
$endgroup$
add a comment |
$begingroup$
No. Right hand limit of $f$ at $x_0$ is $l$ if $f(x_0+r_n) to l$ for every sequence $(r_n)$ decreasing to $0$. A particular sequence will not do. For example, take $f(x)=0$ for $x$ rational and $1$ for $x$ irrational. Take $x_0=0$.
$endgroup$
add a comment |
$begingroup$
No. Right hand limit of $f$ at $x_0$ is $l$ if $f(x_0+r_n) to l$ for every sequence $(r_n)$ decreasing to $0$. A particular sequence will not do. For example, take $f(x)=0$ for $x$ rational and $1$ for $x$ irrational. Take $x_0=0$.
$endgroup$
No. Right hand limit of $f$ at $x_0$ is $l$ if $f(x_0+r_n) to l$ for every sequence $(r_n)$ decreasing to $0$. A particular sequence will not do. For example, take $f(x)=0$ for $x$ rational and $1$ for $x$ irrational. Take $x_0=0$.
answered 46 mins ago
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
76.1k53370
add a comment |
add a comment |
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