I need help with the last half of the proof.

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 $\displaystyle |n-L| >1$