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
edited 48 mins ago
GNUSupporter 8964民主女神 地下教會
14.1k82651
asked 49 mins ago
Math GuyMath Guy
1407
$endgroup$
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
edited 48 mins ago
GNUSupporter 8964民主女神 地下教會
14.1k82651
asked 49 mins ago
Math GuyMath Guy
1407
$endgroup$
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
edited 48 mins ago
GNUSupporter 8964民主女神 地下教會
14.1k82651
asked 49 mins ago
Math GuyMath Guy
1407
$endgroup$
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
asked 49 mins ago
Math GuyMath Guy
1407
edited 48 mins ago
GNUSupporter 8964民主女神 地下教會
14.1k82651
asked 49 mins ago
Math GuyMath Guy
1407
edited 48 mins ago
GNUSupporter 8964民主女神 地下教會
14.1k82651
edited 48 mins ago
GNUSupporter 8964民主女神 地下教會
14.1k82651
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
asked 49 mins ago
Math GuyMath Guy
1407
asked 49 mins ago
Math GuyMath Guy
1407
1407
add a comment |
add a comment |
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, $lim_ntoinftyfleft(frac1nright)=0$.
answered 46 mins ago
José Carlos SantosJosé Carlos Santos
176k24135244
$endgroup$
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 $x_0=0$.
answered 46 mins ago
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
add a comment |
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2 Answers
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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, $lim_ntoinftyfleft(frac1nright)=0$.
answered 46 mins ago
José Carlos SantosJosé Carlos Santos
176k24135244
$endgroup$
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, $lim_ntoinftyfleft(frac1nright)=0$.
answered 46 mins ago
José Carlos SantosJosé Carlos Santos
176k24135244
$endgroup$
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, $lim_ntoinftyfleft(frac1nright)=0$.
answered 46 mins ago
José Carlos SantosJosé Carlos Santos
176k24135244
$endgroup$
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
answered 46 mins ago
José Carlos SantosJosé Carlos Santos
176k24135244
answered 46 mins ago
José Carlos SantosJosé Carlos Santos
176k24135244
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 $x_0=0$.
answered 46 mins ago
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
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 $x_0=0$.
answered 46 mins ago
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
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 $x_0=0$.
answered 46 mins ago
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
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
answered 46 mins ago
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
answered 46 mins ago
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
answered 46 mins ago
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
76.1k53370
add a comment |
add a comment |
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