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


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? ]

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    $\begingroup$ It does exist though. $\endgroup$ – Rebellos Dec 20 '18 at 9:59
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    $\begingroup$ 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
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    $\begingroup$ 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
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What is asserted is that there is no finite limit. Prove by contradiction. Suppose $\lim_{x\to 0+}\frac 1 {x^{2}}=L$ 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.

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