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!
Best Answer
Well, you know that the limit is $\infty$, so it cannot be finite.
Suppose the limit exists (finite) and is $l$. By permanence of sign you know that $l\ge0$.
Let $\varepsilon>0$. Then, by assumption, there exists $\delta>0$ such that, for $0<|x|<\delta$, it holds $$ \left|\frac{1}{x^2}-l\right|<\varepsilon $$ The inequality is the same as $$ |1-lx^2|<\varepsilon x^2 $$ hence $$ -\varepsilon x^2<1-lx^2<\varepsilon x^2 $$ In particular, $(l+\varepsilon)x^2>1$, which is the same as $$ |x|>\sqrt{\frac{1}{l+\varepsilon}} $$ which is a contradiction, because it doesn't hold for $$ x=\frac{1}{2}\min\left(\delta,\sqrt{\frac{1}{l+\varepsilon}}\,\right) $$
You're simply starting wrong: you don't want to find some $\varepsilon>0$; you want to see that it's contradictory to assume that, for every $\varepsilon>0$, …
In a slightly different way. If $$ \lim_{x\to0}f(x)=l $$ then there exists $\delta>0$ such that $f$ is bounded over $(-\delta,\delta)\setminus\{0\}$. But $f(x)=1/x^2$ is unbounded over any punctured neighborhood of $0$.