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Is “$limlimits_n to inftyf(x_0+frac1n)=l$” another way of expressing the right-sided limit?

Limit $mathop lim limits_n to infty fracnleft( a_1…a_n right)^frac1na_1 + … + a_n$Evaluating $lim limits_x to inftyleft(fracf(x_0 + frac 1x)f(x_0 - frac 1x)right)^x$How 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 $Prove $limlimits_x to +infty fracf(x)x = limlimits_x to +infty f'(x)$ if both limits existDoes, $mathop lim limits_x to +infty f'(x) = + infty Leftrightarrow mathop lim limits_x to +infty fracf(x)x = + infty $?Formal limit definition in $limlimits_xto+infty fracx^2+29x=+infty$How can I prove that $limlimits_x to x_0 left|fracf(x)g(x)right| = infty$ when working with punctured neighborhoods?Calculate the limit $lim limits_ x to infty left(fracx^2+1x-1right)$

0

$begingroup$

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

share|cite|improve this question

edited Apr 19 at 12:53

user21820

40.4k544163

asked Apr 19 at 10:24

Math GuyMath Guy

1657

$endgroup$

add a comment | 

0

$begingroup$

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

share|cite|improve this question

edited Apr 19 at 12:53

user21820

40.4k544163

asked Apr 19 at 10:24

Math GuyMath Guy

1657

$endgroup$

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

share|cite|improve this question

edited Apr 19 at 12:53

user21820

40.4k544163

asked Apr 19 at 10:24

Math GuyMath Guy

1657

$endgroup$

share|cite|improve this question

edited Apr 19 at 12:53

user21820

40.4k544163

asked Apr 19 at 10:24

Math GuyMath Guy

1657

share|cite|improve this question

edited Apr 19 at 12:53

user21820

40.4k544163

asked Apr 19 at 10:24

Math GuyMath Guy

1657

edited Apr 19 at 12:53

user21820

40.4k544163

edited Apr 19 at 12:53

user21820

40.4k544163

asked Apr 19 at 10:24

Math GuyMath Guy

1657

asked Apr 19 at 10:24

Math GuyMath Guy

1657

2 Answers
2

active

oldest

votes

10

$begingroup$

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, in spite of the fact that $lim_ntoinftyfleft(frac1nright)=0$.

share|cite|improve this answer

edited Apr 19 at 20:51

answered Apr 19 at 10:27

José Carlos SantosJosé Carlos Santos

178k24139251

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Who said that $n$ was an integer ? ;-)
  $endgroup$
  – Yves Daoust
  Apr 19 at 10:35
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 19 at 10:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Apr 19 at 11:42
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
  $endgroup$
  – Yves Daoust
  Apr 19 at 11:47
  
  

  
  
  
  

add a comment | 

6

$begingroup$

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 $x_0=0$.

share|cite|improve this answer

answered Apr 19 at 10:27

Kavi Rama MurthyKavi Rama Murthy

78.2k53571

$endgroup$

add a comment | 

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10

$begingroup$

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, in spite of the fact that $lim_ntoinftyfleft(frac1nright)=0$.

share|cite|improve this answer

edited Apr 19 at 20:51

answered Apr 19 at 10:27

José Carlos SantosJosé Carlos Santos

178k24139251

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Who said that $n$ was an integer ? ;-)
  $endgroup$
  – Yves Daoust
  Apr 19 at 10:35
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 19 at 10:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Apr 19 at 11:42
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
  $endgroup$
  – Yves Daoust
  Apr 19 at 11:47
  
  

  
  
  
  

add a comment | 

10

$begingroup$

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, in spite of the fact that $lim_ntoinftyfleft(frac1nright)=0$.

share|cite|improve this answer

edited Apr 19 at 20:51

answered Apr 19 at 10:27

José Carlos SantosJosé Carlos Santos

178k24139251

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Who said that $n$ was an integer ? ;-)
  $endgroup$
  – Yves Daoust
  Apr 19 at 10:35
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 19 at 10:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Apr 19 at 11:42
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
  $endgroup$
  – Yves Daoust
  Apr 19 at 11:47
  
  

  
  
  
  

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, in spite of the fact that $lim_ntoinftyfleft(frac1nright)=0$.

share|cite|improve this answer

edited Apr 19 at 20:51

answered Apr 19 at 10:27

José Carlos SantosJosé Carlos Santos

178k24139251

$endgroup$

share|cite|improve this answer

edited Apr 19 at 20:51

answered Apr 19 at 10:27

José Carlos SantosJosé Carlos Santos

178k24139251

answered Apr 19 at 10:27

José Carlos SantosJosé Carlos Santos

178k24139251

answered Apr 19 at 10:27

José Carlos SantosJosé Carlos Santos

178k24139251

6

$begingroup$

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 $x_0=0$.

share|cite|improve this answer

answered Apr 19 at 10:27

Kavi Rama MurthyKavi Rama Murthy

78.2k53571

$endgroup$

add a comment | 

6

$begingroup$

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 $x_0=0$.

share|cite|improve this answer

answered Apr 19 at 10:27

Kavi Rama MurthyKavi Rama Murthy

78.2k53571

$endgroup$

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 $x_0=0$.

share|cite|improve this answer

answered Apr 19 at 10:27

Kavi Rama MurthyKavi Rama Murthy

78.2k53571

$endgroup$

share|cite|improve this answer

answered Apr 19 at 10:27

Kavi Rama MurthyKavi Rama Murthy

78.2k53571

answered Apr 19 at 10:27

Kavi Rama MurthyKavi Rama Murthy

78.2k53571

answered Apr 19 at 10:27

Kavi Rama MurthyKavi Rama Murthy

78.2k53571