show that the limit does not exist: $lim_{xto0}frac{1}{x^2}$












1












$begingroup$



Show that the following limit does not exist:
$$lim_{xto 0}frac{1}{x^2}$$
$(x>0)$





The $delta$ - $varepsilon$ definition can be used to prove a given limit exists for some function at a particular point. My question is, can we prove the non-existence of a limit using the $delta$ - $varepsilon$ definition? (A little hint on how, if yes.)

Besides that, what (other) methods can we use?




[SIMILAR POST: Can we prove that there is no limit at $x=0$ for $f(x)=1/x$ using epsilon-delta definition? ]






real-analysis limits







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  • 2




    $begingroup$
    It does exist though.
    $endgroup$
    – Rebellos
    Dec 20 '18 at 9:59






  • 1




    $begingroup$
    The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
    $endgroup$
    – Daniele Tampieri
    Dec 20 '18 at 10:00






  • 1




    $begingroup$
    Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 10:00


















1












$begingroup$



Show that the following limit does not exist:
$$lim_{xto 0}frac{1}{x^2}$$
$(x>0)$





The $delta$ - $varepsilon$ definition can be used to prove a given limit exists for some function at a particular point. My question is, can we prove the non-existence of a limit using the $delta$ - $varepsilon$ definition? (A little hint on how, if yes.)

Besides that, what (other) methods can we use?




[SIMILAR POST: Can we prove that there is no limit at $x=0$ for $f(x)=1/x$ using epsilon-delta definition? ]






real-analysis limits







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It does exist though.
    $endgroup$
    – Rebellos
    Dec 20 '18 at 9:59






  • 1




    $begingroup$
    The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
    $endgroup$
    – Daniele Tampieri
    Dec 20 '18 at 10:00






  • 1




    $begingroup$
    Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 10:00
















1












1








1





$begingroup$



Show that the following limit does not exist:
$$lim_{xto 0}frac{1}{x^2}$$
$(x>0)$





The $delta$ - $varepsilon$ definition can be used to prove a given limit exists for some function at a particular point. My question is, can we prove the non-existence of a limit using the $delta$ - $varepsilon$ definition? (A little hint on how, if yes.)

Besides that, what (other) methods can we use?




[SIMILAR POST: Can we prove that there is no limit at $x=0$ for $f(x)=1/x$ using epsilon-delta definition? ]






real-analysis limits







share|cite|improve this question











$endgroup$





Show that the following limit does not exist:
$$lim_{xto 0}frac{1}{x^2}$$
$(x>0)$





The $delta$ - $varepsilon$ definition can be used to prove a given limit exists for some function at a particular point. My question is, can we prove the non-existence of a limit using the $delta$ - $varepsilon$ definition? (A little hint on how, if yes.)

Besides that, what (other) methods can we use?




[SIMILAR POST: Can we prove that there is no limit at $x=0$ for $f(x)=1/x$ using epsilon-delta definition? ]






real-analysis limits




real-analysis limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 10:14







Za Ira

















asked Dec 20 '18 at 9:51









Za IraZa Ira

161115




161115








  • 2




    $begingroup$
    It does exist though.
    $endgroup$
    – Rebellos
    Dec 20 '18 at 9:59






  • 1




    $begingroup$
    The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
    $endgroup$
    – Daniele Tampieri
    Dec 20 '18 at 10:00






  • 1




    $begingroup$
    Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 10:00
















  • 2




    $begingroup$
    It does exist though.
    $endgroup$
    – Rebellos
    Dec 20 '18 at 9:59






  • 1




    $begingroup$
    The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
    $endgroup$
    – Daniele Tampieri
    Dec 20 '18 at 10:00






  • 1




    $begingroup$
    Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 10:00










2




2




$begingroup$
It does exist though.
$endgroup$
– Rebellos
Dec 20 '18 at 9:59




$begingroup$
It does exist though.
$endgroup$
– Rebellos
Dec 20 '18 at 9:59




1




1




$begingroup$
The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
$endgroup$
– Daniele Tampieri
Dec 20 '18 at 10:00




$begingroup$
The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
$endgroup$
– Daniele Tampieri
Dec 20 '18 at 10:00




1




1




$begingroup$
Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:00






$begingroup$
Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:00












1 Answer
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$begingroup$

What is asserted is that there is no finite limit. Prove by contradiction. Suppose $lim_{xto 0+}frac 1 {x^{2}}=L$ exists and is finite. Then we can find $delta >0$ such that $|frac 1 {x^{2}}-L|<1$ whenever $0<x<delta$. If $n$ is any sufficiently large integer then $0<frac 1 n<delta$ so, taking $x=frac 1 n$ we get $| n^{2}-L|<1$. This gives $n^{2} <L+1$ for any sufficiently large integer $n$ which is obviously a contradiction.






share|cite|improve this answer









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    $begingroup$

    What is asserted is that there is no finite limit. Prove by contradiction. Suppose $lim_{xto 0+}frac 1 {x^{2}}=L$ exists and is finite. Then we can find $delta >0$ such that $|frac 1 {x^{2}}-L|<1$ whenever $0<x<delta$. If $n$ is any sufficiently large integer then $0<frac 1 n<delta$ so, taking $x=frac 1 n$ we get $| n^{2}-L|<1$. This gives $n^{2} <L+1$ for any sufficiently large integer $n$ which is obviously a contradiction.






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      What is asserted is that there is no finite limit. Prove by contradiction. Suppose $lim_{xto 0+}frac 1 {x^{2}}=L$ exists and is finite. Then we can find $delta >0$ such that $|frac 1 {x^{2}}-L|<1$ whenever $0<x<delta$. If $n$ is any sufficiently large integer then $0<frac 1 n<delta$ so, taking $x=frac 1 n$ we get $| n^{2}-L|<1$. This gives $n^{2} <L+1$ for any sufficiently large integer $n$ which is obviously a contradiction.






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        What is asserted is that there is no finite limit. Prove by contradiction. Suppose $lim_{xto 0+}frac 1 {x^{2}}=L$ exists and is finite. Then we can find $delta >0$ such that $|frac 1 {x^{2}}-L|<1$ whenever $0<x<delta$. If $n$ is any sufficiently large integer then $0<frac 1 n<delta$ so, taking $x=frac 1 n$ we get $| n^{2}-L|<1$. This gives $n^{2} <L+1$ for any sufficiently large integer $n$ which is obviously a contradiction.






        share|cite|improve this answer









        $endgroup$



        What is asserted is that there is no finite limit. Prove by contradiction. Suppose $lim_{xto 0+}frac 1 {x^{2}}=L$ exists and is finite. Then we can find $delta >0$ such that $|frac 1 {x^{2}}-L|<1$ whenever $0<x<delta$. If $n$ is any sufficiently large integer then $0<frac 1 n<delta$ so, taking $x=frac 1 n$ we get $| n^{2}-L|<1$. This gives $n^{2} <L+1$ for any sufficiently large integer $n$ which is obviously a contradiction.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 10:00









        Kavi Rama MurthyKavi Rama Murthy

        60.6k42161




        60.6k42161






























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