Is “$limlimits_n to inftyf(x_0+frac1n)=l$” another way of expressing the right-sided limit? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)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 $lim limits_ x to infty left(fracx^2+1x-1right)$

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



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)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 $lim limits_ x to infty left(fracx^2+1x-1right)$










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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











$endgroup$
















    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











    $endgroup$














      0












      0








      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











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      user21820

      40.4k544163




      40.4k544163










      asked 2 days ago









      Math GuyMath Guy

      1607




      1607




















          2 Answers
          2






          active

          oldest

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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











          $endgroup$








          • 2




            $begingroup$
            Who said that $n$ was an integer ? ;-)
            $endgroup$
            – Yves Daoust
            2 days ago






          • 2




            $begingroup$
            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
            2 days 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
            2 days ago







          • 2




            $begingroup$
            @MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
            $endgroup$
            – Yves Daoust
            2 days ago



















          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









          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            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











            $endgroup$








            • 2




              $begingroup$
              Who said that $n$ was an integer ? ;-)
              $endgroup$
              – Yves Daoust
              2 days ago






            • 2




              $begingroup$
              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
              2 days 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
              2 days ago







            • 2




              $begingroup$
              @MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
              $endgroup$
              – Yves Daoust
              2 days ago
















            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











            $endgroup$








            • 2




              $begingroup$
              Who said that $n$ was an integer ? ;-)
              $endgroup$
              – Yves Daoust
              2 days ago






            • 2




              $begingroup$
              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
              2 days 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
              2 days ago







            • 2




              $begingroup$
              @MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
              $endgroup$
              – Yves Daoust
              2 days ago














            10












            10








            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











            $endgroup$



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







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered 2 days ago









            José Carlos SantosJosé Carlos Santos

            176k24137247




            176k24137247







            • 2




              $begingroup$
              Who said that $n$ was an integer ? ;-)
              $endgroup$
              – Yves Daoust
              2 days ago






            • 2




              $begingroup$
              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
              2 days 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
              2 days ago







            • 2




              $begingroup$
              @MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
              $endgroup$
              – Yves Daoust
              2 days ago













            • 2




              $begingroup$
              Who said that $n$ was an integer ? ;-)
              $endgroup$
              – Yves Daoust
              2 days ago






            • 2




              $begingroup$
              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
              2 days 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
              2 days ago







            • 2




              $begingroup$
              @MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
              $endgroup$
              – Yves Daoust
              2 days ago








            2




            2




            $begingroup$
            Who said that $n$ was an integer ? ;-)
            $endgroup$
            – Yves Daoust
            2 days ago




            $begingroup$
            Who said that $n$ was an integer ? ;-)
            $endgroup$
            – Yves Daoust
            2 days ago




            2




            2




            $begingroup$
            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
            2 days 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
            2 days 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
            2 days 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
            2 days ago





            2




            2




            $begingroup$
            @MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
            $endgroup$
            – Yves Daoust
            2 days ago





            $begingroup$
            @MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
            $endgroup$
            – Yves Daoust
            2 days ago












            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 2 days ago









                Kavi Rama MurthyKavi Rama Murthy

                76.7k53471




                76.7k53471



























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