Is “$limlimits_{n to infty}f(x_0+frac{1}{n})=l$” another way of expressing the right-sided limit?
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Let $f:mathbb{R} to mathbb{R}$. Can we say that $limlimits_{n to infty}f(x_0+frac{1}{n})=l$ 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
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add a comment |
$begingroup$
Let $f:mathbb{R} to mathbb{R}$. Can we say that $limlimits_{n to infty}f(x_0+frac{1}{n})=l$ 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 $f:mathbb{R} to mathbb{R}$. Can we say that $limlimits_{n to infty}f(x_0+frac{1}{n})=l$ 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 $f:mathbb{R} to mathbb{R}$. Can we say that $limlimits_{n to infty}f(x_0+frac{1}{n})=l$ 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 15 hours ago
user21820
40.4k544163
40.4k544163
asked 17 hours ago
Math GuyMath Guy
1507
1507
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2 Answers
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No, it is not true. Take, for instance,$$f(x)=begin{cases}sinleft(fracpi xright)&text{ if }xneq0\0&text{ if }x=0.end{cases}$$Then the limit $lim_{xto0^+}f(x)$ doesn't exist, in spite of the fact that $lim_{ntoinfty}fleft(frac1nright)=0$.
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1
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Who said that $n$ was an integer ? ;-)
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– Yves Daoust
17 hours ago
1
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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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. So, there is no ambiguity.
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– José Carlos Santos
17 hours ago
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@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
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– Math Guy
16 hours ago
1
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@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
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– Yves Daoust
16 hours ago
add a comment |
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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 $x_0=0$.
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2 Answers
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$begingroup$
No, it is not true. Take, for instance,$$f(x)=begin{cases}sinleft(fracpi xright)&text{ if }xneq0\0&text{ if }x=0.end{cases}$$Then the limit $lim_{xto0^+}f(x)$ doesn't exist, in spite of the fact that $lim_{ntoinfty}fleft(frac1nright)=0$.
$endgroup$
1
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
17 hours 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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
17 hours ago
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
16 hours ago
1
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
16 hours ago
add a comment |
$begingroup$
No, it is not true. Take, for instance,$$f(x)=begin{cases}sinleft(fracpi xright)&text{ if }xneq0\0&text{ if }x=0.end{cases}$$Then the limit $lim_{xto0^+}f(x)$ doesn't exist, in spite of the fact that $lim_{ntoinfty}fleft(frac1nright)=0$.
$endgroup$
1
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
17 hours 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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
17 hours ago
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
16 hours ago
1
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
16 hours ago
add a comment |
$begingroup$
No, it is not true. Take, for instance,$$f(x)=begin{cases}sinleft(fracpi xright)&text{ if }xneq0\0&text{ if }x=0.end{cases}$$Then the limit $lim_{xto0^+}f(x)$ doesn't exist, in spite of the fact that $lim_{ntoinfty}fleft(frac1nright)=0$.
$endgroup$
No, it is not true. Take, for instance,$$f(x)=begin{cases}sinleft(fracpi xright)&text{ if }xneq0\0&text{ if }x=0.end{cases}$$Then the limit $lim_{xto0^+}f(x)$ doesn't exist, in spite of the fact that $lim_{ntoinfty}fleft(frac1nright)=0$.
edited 7 hours ago
answered 17 hours ago
José Carlos SantosJosé Carlos Santos
176k24136245
176k24136245
1
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
17 hours 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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
17 hours ago
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
16 hours ago
1
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
16 hours ago
add a comment |
1
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
17 hours 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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
17 hours ago
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
16 hours ago
1
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
16 hours ago
1
1
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
17 hours ago
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
17 hours 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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
17 hours 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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
17 hours ago
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
16 hours ago
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
16 hours ago
1
1
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
16 hours ago
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
16 hours 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 17 hours ago
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
76.1k53370
add a comment |
add a comment |
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