# show that the limit does not exist: $\lim_{x\to0}\frac{1}{x^2}$

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? ]

• It does exist though. – Rebellos Dec 20 '18 at 9:59
• 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 $\infty$. – Daniele Tampieri Dec 20 '18 at 10:00
• Show that for every $L>0$ and $\varepsilon=1\;\exists\:\delta>0$, $|f(x)-L|>\varepsilon$ whenever $|x|<\delta$. – Yadati Kiran Dec 20 '18 at 10:00

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 $$0. If $$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} for any sufficiently large integer $$n$$ which is obviously a contradiction.