Proof that $\lim\limits_{x \to 0} \frac{1}{x^{2}}$ does not exist I want to come up with a good argument as to why $\lim\limits_{x \to 0} \frac{1}{x^{2}}$ DNE. Here is my attempt.
Proof by Contradiction: Suppose $\lim\limits_{x \to 0} \frac{1}{x^{2}}=L$ where $L\in(-\infty,\infty)$.
Well for any $x\in(-\infty,-1)\cup(1,\infty)$, $0<\frac{1}{x^{2}}<1$.
$\lim\limits_{x \to 0} \frac{1}{x^{2}}=L\implies0<\frac{1}{x^{2}}\leq L$ for any $x\in[-1,0)\cup(0,1]$.
So, $0<\frac{1}{x^{2}}<1+L$ for any $x\in(-\infty,0)\cup(0,\infty)$.
Let's consider $x=\frac{1}{\sqrt{1+L}}\in(-\infty,0)\cup(0,\infty)$ $[0<1+L]$.
Then,
$\frac{1}{\big(\frac{1}{\sqrt{1+L}}\big)^{2}}<1+L\implies(\sqrt{1+L})^{2}<1+L\implies|1+L|<1+L$
$\implies 1+L<1+L$.
A contradiction. Thus, $\lim\limits_{x \to 0} \frac{1}{x^{2}}$ DNE.
Any comment would be appreciated.
 A: 
$\lim\limits_{x \to 0} \frac{1}{x^{2}}=L\implies0<\frac{1}{x^{2}}\leq L$ for any $x\in[-1,0) \cup (0,1]$.

This line is incorrect.  Say $-h^2 \to 0 \ge 0$ as $h \to 0$, but $-h^2 \le 0$.
Limits preserves monotonicity.  That is, if $f(x) \ge 0$ on an open interval containing a point $a$, and $\lim_{x\to a}f(x) = L \in \Bbb{R}$, then $L \ge 0$.
However, the converse is incorrect.  I've given a counterexample above.
It's nice to start with a proof by contradiction, but I suggest that you use the definition ($\epsilon$-$\delta$ criterion) of the limit with $\epsilon = 1$ to establish a $\delta$-neighbourhood $(-\delta,\delta)$ about $x = 0$ so that for all $0 < |x|<\delta$, $$\left|\frac{1}{x^2} - L\right| < \epsilon = 1.$$
$$ L - 1 < \frac{1}{x^2} < L + 1. $$
In particular, $$x > \frac{1}{\sqrt{L+1}} \label{wrong1} \tag{$\star$}$$
Here, both $\delta$ and $L$ exist, and they come from your hypothesis (to be rejected by contradition later).
Note that the quantifier in front of "$0 < |x| < \delta$" is "for all", so you can choose some small $x_0\in (-\delta,\delta) \setminus \{0\}$ so that \eqref{wrong1} is violated, say $x_0 = \min\left\{\frac\delta2, \frac{1}{2\sqrt{L+1}}\right\}$.
This proves that the limit $L$ doesn't exists.
A: I prefer to prove the claim by contradiction, but with no $\varepsilon-\delta$ analysis, which I use when I am really desperate.
Assume $$\lim_{x\to 0}{1\over x^2}=L$$ Then
$$1=\lim_{x\to 0}x^2{1\over x^2}=\lim_{x\to 0}x^2\lim_{x\to 0}{1\over x^2}=0\cdot L=0$$
