Proving $lim_xto 0 dfrac1 + 1/x^21 + 1/x$ does not exist using the sequence definition of a limit

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I'm working through problems in a real analysis textbook. I don't have any solutions. I am using the following definition for a limit:



Definition: Given a function $f : D rightarrow mathbbR$ and a limit point $x_0 in D$, for a number $ell$, we write



$$lim_xto x_0 f(x) = ell $$



provided that whenever $x_n in D - x_0$ that converges to $x_0$,



$$lim_ntoinfty f(x_n) = ell.$$



Problem:



I am trying to show that the limit



$$lim_xto 0 dfrac1 + 1/x^21 + 1/x$$



does not exist.



My attempt:



Let $x_n$ be a sequence in $D - 0$ that converges to $0$. Now we need to show that



$$lim_nto infty dfrac1 + 1/x_n^21 + 1/x_n = ell.$$



Rewrite the limit as follows:



$$ lim_nto inftydfrac1 + 1/x_n^21 + 1/x_n = lim_nto infty dfracx_n^2 + 1x_n^2 + x_n = dfrac10,$$



which is undefined. Thus, we can conclude that our limit does not exist.







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  • 2




    That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
    – user582578
    Aug 8 at 20:20















up vote
1
down vote

favorite












I'm working through problems in a real analysis textbook. I don't have any solutions. I am using the following definition for a limit:



Definition: Given a function $f : D rightarrow mathbbR$ and a limit point $x_0 in D$, for a number $ell$, we write



$$lim_xto x_0 f(x) = ell $$



provided that whenever $x_n in D - x_0$ that converges to $x_0$,



$$lim_ntoinfty f(x_n) = ell.$$



Problem:



I am trying to show that the limit



$$lim_xto 0 dfrac1 + 1/x^21 + 1/x$$



does not exist.



My attempt:



Let $x_n$ be a sequence in $D - 0$ that converges to $0$. Now we need to show that



$$lim_nto infty dfrac1 + 1/x_n^21 + 1/x_n = ell.$$



Rewrite the limit as follows:



$$ lim_nto inftydfrac1 + 1/x_n^21 + 1/x_n = lim_nto infty dfracx_n^2 + 1x_n^2 + x_n = dfrac10,$$



which is undefined. Thus, we can conclude that our limit does not exist.







share|cite|improve this question
















  • 2




    That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
    – user582578
    Aug 8 at 20:20













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I'm working through problems in a real analysis textbook. I don't have any solutions. I am using the following definition for a limit:



Definition: Given a function $f : D rightarrow mathbbR$ and a limit point $x_0 in D$, for a number $ell$, we write



$$lim_xto x_0 f(x) = ell $$



provided that whenever $x_n in D - x_0$ that converges to $x_0$,



$$lim_ntoinfty f(x_n) = ell.$$



Problem:



I am trying to show that the limit



$$lim_xto 0 dfrac1 + 1/x^21 + 1/x$$



does not exist.



My attempt:



Let $x_n$ be a sequence in $D - 0$ that converges to $0$. Now we need to show that



$$lim_nto infty dfrac1 + 1/x_n^21 + 1/x_n = ell.$$



Rewrite the limit as follows:



$$ lim_nto inftydfrac1 + 1/x_n^21 + 1/x_n = lim_nto infty dfracx_n^2 + 1x_n^2 + x_n = dfrac10,$$



which is undefined. Thus, we can conclude that our limit does not exist.







share|cite|improve this question












I'm working through problems in a real analysis textbook. I don't have any solutions. I am using the following definition for a limit:



Definition: Given a function $f : D rightarrow mathbbR$ and a limit point $x_0 in D$, for a number $ell$, we write



$$lim_xto x_0 f(x) = ell $$



provided that whenever $x_n in D - x_0$ that converges to $x_0$,



$$lim_ntoinfty f(x_n) = ell.$$



Problem:



I am trying to show that the limit



$$lim_xto 0 dfrac1 + 1/x^21 + 1/x$$



does not exist.



My attempt:



Let $x_n$ be a sequence in $D - 0$ that converges to $0$. Now we need to show that



$$lim_nto infty dfrac1 + 1/x_n^21 + 1/x_n = ell.$$



Rewrite the limit as follows:



$$ lim_nto inftydfrac1 + 1/x_n^21 + 1/x_n = lim_nto infty dfracx_n^2 + 1x_n^2 + x_n = dfrac10,$$



which is undefined. Thus, we can conclude that our limit does not exist.









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share|cite|improve this question




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asked Aug 8 at 20:06









Hat

789115




789115







  • 2




    That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
    – user582578
    Aug 8 at 20:20













  • 2




    That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
    – user582578
    Aug 8 at 20:20








2




2




That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
– user582578
Aug 8 at 20:20





That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
– user582578
Aug 8 at 20:20











2 Answers
2






active

oldest

votes

















up vote
2
down vote













As I know, in order to show in this way the limit doesn't exist we
have to address to a specific sequence. Take for example $a_n=frac1n$
which converges to $0$, but



beginalign*
fleft(a_nright) & =fleft(frac1nright)=frac1+frac1left(frac1nright)^21+frac1left(frac1nright)=frac1+n^21+n=fracleft(1+nright)^2-2n1+n=1+n-frac2n1+n=\
& =1+n-frac2left(1+nright)-21+n=-1+n+frac21+ntoinfty
endalign*






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  • 1




    You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
    – user582578
    Aug 8 at 20:44

















up vote
0
down vote













The sequence $x_n=1/n$ converges to $0$; on the other hand,
$$
f(1/n)=frac1+n^21+n>fracn^22n=fracn2
$$



Therefore, the sequence $f(1/n)$ doesn't converge.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote













    As I know, in order to show in this way the limit doesn't exist we
    have to address to a specific sequence. Take for example $a_n=frac1n$
    which converges to $0$, but



    beginalign*
    fleft(a_nright) & =fleft(frac1nright)=frac1+frac1left(frac1nright)^21+frac1left(frac1nright)=frac1+n^21+n=fracleft(1+nright)^2-2n1+n=1+n-frac2n1+n=\
    & =1+n-frac2left(1+nright)-21+n=-1+n+frac21+ntoinfty
    endalign*






    share|cite|improve this answer
















    • 1




      You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
      – user582578
      Aug 8 at 20:44














    up vote
    2
    down vote













    As I know, in order to show in this way the limit doesn't exist we
    have to address to a specific sequence. Take for example $a_n=frac1n$
    which converges to $0$, but



    beginalign*
    fleft(a_nright) & =fleft(frac1nright)=frac1+frac1left(frac1nright)^21+frac1left(frac1nright)=frac1+n^21+n=fracleft(1+nright)^2-2n1+n=1+n-frac2n1+n=\
    & =1+n-frac2left(1+nright)-21+n=-1+n+frac21+ntoinfty
    endalign*






    share|cite|improve this answer
















    • 1




      You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
      – user582578
      Aug 8 at 20:44












    up vote
    2
    down vote










    up vote
    2
    down vote









    As I know, in order to show in this way the limit doesn't exist we
    have to address to a specific sequence. Take for example $a_n=frac1n$
    which converges to $0$, but



    beginalign*
    fleft(a_nright) & =fleft(frac1nright)=frac1+frac1left(frac1nright)^21+frac1left(frac1nright)=frac1+n^21+n=fracleft(1+nright)^2-2n1+n=1+n-frac2n1+n=\
    & =1+n-frac2left(1+nright)-21+n=-1+n+frac21+ntoinfty
    endalign*






    share|cite|improve this answer












    As I know, in order to show in this way the limit doesn't exist we
    have to address to a specific sequence. Take for example $a_n=frac1n$
    which converges to $0$, but



    beginalign*
    fleft(a_nright) & =fleft(frac1nright)=frac1+frac1left(frac1nright)^21+frac1left(frac1nright)=frac1+n^21+n=fracleft(1+nright)^2-2n1+n=1+n-frac2n1+n=\
    & =1+n-frac2left(1+nright)-21+n=-1+n+frac21+ntoinfty
    endalign*







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 8 at 20:35









    Jon

    500413




    500413







    • 1




      You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
      – user582578
      Aug 8 at 20:44












    • 1




      You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
      – user582578
      Aug 8 at 20:44







    1




    1




    You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
    – user582578
    Aug 8 at 20:44




    You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
    – user582578
    Aug 8 at 20:44










    up vote
    0
    down vote













    The sequence $x_n=1/n$ converges to $0$; on the other hand,
    $$
    f(1/n)=frac1+n^21+n>fracn^22n=fracn2
    $$



    Therefore, the sequence $f(1/n)$ doesn't converge.






    share|cite|improve this answer
























      up vote
      0
      down vote













      The sequence $x_n=1/n$ converges to $0$; on the other hand,
      $$
      f(1/n)=frac1+n^21+n>fracn^22n=fracn2
      $$



      Therefore, the sequence $f(1/n)$ doesn't converge.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        The sequence $x_n=1/n$ converges to $0$; on the other hand,
        $$
        f(1/n)=frac1+n^21+n>fracn^22n=fracn2
        $$



        Therefore, the sequence $f(1/n)$ doesn't converge.






        share|cite|improve this answer












        The sequence $x_n=1/n$ converges to $0$; on the other hand,
        $$
        f(1/n)=frac1+n^21+n>fracn^22n=fracn2
        $$



        Therefore, the sequence $f(1/n)$ doesn't converge.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 8 at 20:50









        egreg

        165k1180187




        165k1180187






















             

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