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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    $begingroup$


    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







    share|cite|improve this question











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      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






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      edited Apr 19 at 12:53









      user21820

      40.4k544163




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      asked Apr 19 at 10:24









      Math GuyMath Guy

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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 $lim_{ntoinfty}fleft(frac1nright)=0$.






          share|cite|improve this answer











          $endgroup$









          • 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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. 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





















          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 $x_0=0$.






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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

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            10












            $begingroup$

            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 $lim_{ntoinfty}fleft(frac1nright)=0$.






            share|cite|improve this answer











            $endgroup$









            • 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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. 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


















            10












            $begingroup$

            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 $lim_{ntoinfty}fleft(frac1nright)=0$.






            share|cite|improve this answer











            $endgroup$









            • 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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. 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
















            10












            10








            10





            $begingroup$

            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 $lim_{ntoinfty}fleft(frac1nright)=0$.






            share|cite|improve this answer











            $endgroup$



            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 $lim_{ntoinfty}fleft(frac1nright)=0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 19 at 20:51

























            answered Apr 19 at 10:27









            José Carlos SantosJosé Carlos Santos

            177k24138250




            177k24138250








            • 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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. 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
















            • 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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. 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










            2




            2




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




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




            3




            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 $lim_{ninmathbb N}a_n$, instead of $lim_{ntoinfty}a_n$. So, there is no ambiguity.
            $endgroup$
            – José Carlos Santos
            Apr 19 at 10:39




            $begingroup$
            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
            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






            $begingroup$
            @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




            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






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













            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









            $endgroup$


















              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









              $endgroup$
















                6












                6








                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









                $endgroup$



                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 $x_0=0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Apr 19 at 10:27









                Kavi Rama MurthyKavi Rama Murthy

                77.3k53471




                77.3k53471






























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