Let $n \in \mathbb{N}$ be such that $n \geq 3$. Derive the inequality $-x^2 \leq x^n \leq x^2$ for $-1 < x < 1$. Then, use the fact that $\lim_{x \to 0} x^2 = 0$ to show that $\lim_{x \to 0} x^n = 0$.
I suppose there are two questions here. Firstly, I did not see how to prove that $-x^2 \leq x^n \leq x^2$ for $-1 < x < 1$. This is one of those things I find to be extremely "obvious" that it ends up being difficult to see how to prove it! A hint to get started on this would be helpful!
I decided to attempt to prove the final statement without using the previous, and I did come up with something. I would like to confirm this since ignoring part of a question is usually a bad idea!
I want to show $\forall \epsilon > 0$, $\exists \delta > 0$ s.t. whenever $|x| < \delta$, $|x^n|< \epsilon$.
Since $x \in (-1,1)$, I only need to consider $(0,1)$ since we are taking the absolute value. Write $x = \frac{1}{1 + y}$ for some $y > 0$.
Choose $\delta = min(1,(1-n)\epsilon)$. Then,
$|x^n| = |\frac{x}{(1 + y)^{n-1}}| < |\frac{\delta}{1 - n + ny}| < |\frac{\delta}{1-n}| = \epsilon$
where the middle step followed by Bernoulli's inequality. As I said, I weary having skipped an entire portion of the problem, but I am curious if this works!