Prove that \(\displaystyle \lim_{x \to 0}\frac{1}{x^2}\) does not exist.

Proof:

Assume to the contrary that \(\displaystyle \lim_{x \to 0}\frac{1}{x^2}\) this exists. Then there should exists a real number \(\displaystyle L\) such that \(\displaystyle \lim_{x \to 0}\frac{1}{x^2}=L\). Let \(\displaystyle \epsilon =1\). Then there exist \(\displaystyle \delta >0\). If \(\displaystyle x\) is a real number for which \(\displaystyle 0<|x|<\delta\), then \(\displaystyle |\frac{1}{x^2}-L|<\epsilon=1\). Choose an integer \(\displaystyle n >\lceil 1/\delta \rceil \geq 1\). Since \(\displaystyle n>1/\delta\), it follows that \(\displaystyle 0<1/n<\delta\).

Let \(\displaystyle x=1/n\). I want to generate contradiction by showing that

\(\displaystyle |\frac{1}{x^2}-L|=|n^2-L|>1\)

With \(\displaystyle n>1\) and \(\displaystyle L\geq 0\), how can show that \(\displaystyle |n-L| >1\)