Show that $\lim_{x \to 0} \frac{1}{x^2}$ does not exist.
I have seen this: Epsilon-delta proof that $ \lim_{x\to 0} {1\over x^2}$ does not exist but there is a step in the answer that I am not fully grasping. I wanted to try my own argument and see if I could convince myself.
In my argument, I use a step I am not sure I am allowed to use. If my method is simply not going to work, please let me know and I will modify the question to ask for clarification on the other answer. Also (if possible) I would prefer a hint than an answer unless I end up completely stuck!
I want to show $\exists \epsilon > 0 \forall \delta > 0$ s.t. whenever $|x| < \delta$, $|\frac{1}{x^2} - L| \geq \epsilon$
Side work: $x < \delta$
$x^2 < \delta^2$
$-x^2 > -\delta^2$ and $\frac{1}{\delta^2} < \frac{1}{x^2}$
$-Lx^2 > -L \delta^2$
$1 - Lx^2 > 1 - L \delta^2$
With this being done, choose $\epsilon = \frac{1 - L \delta^2}{\delta^2}$. This is the part I am unsure of. I have to show this is true $\forall \epsilon > 0$, and because there is a $\delta$ in both the numerator and denominator, I'm not sure this choice of $\epsilon$ is arbitrarily small.
If I continue I would have
$| \frac{1}{x^2} - \frac{Lx^2}{x^2}| = |\frac{1 - L^2x^2}{x^2}| > |\frac{1 - L^2\delta^2}{\delta^2}| = \epsilon$
The only issue is whether or not I can use that $\epsilon$. I am leaning towards no because if I break this back up into two fractions:
$|\frac{1}{\delta^2} - \frac{L\delta^2}{\delta^2}| \geq \frac{1}{\delta^2} - L$, and the first fraction becomes arbitrarily large due to $\delta$ being arbitrarily small in the denominator.
Is there a way I can modify my method to get it to work? Or is this doomed to fail?
Thanks in advance!