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? ]
Best Answer
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.