Can't figure out easy limit: $\lim_{x\to 0^+}\frac{\sin(2x^2)}{x^3}$

Question

I was trying to help someone with their intro calc homework but got stumped by this super easy problem. I know we can apply L'Hôpital's rule to get $\lim_{x\to 0^+}\frac{4\cos(2x^2)}{3x}$ which goes to $\frac{4}{0}= + \infty$ but I think this doesn't count as a proof. What am I forgetting?

Answer 1

You can rewrite the limit as $$\lim_{x\to 0^+} \frac{2}{x}\cdot\frac{\sin 2x^2}{2x^2}$$

and, since $$\frac{\sin 2x^2}{2x^2}$$ is bounded and positive, the limit is $\infty$

Answer 2

You did not forget nothing. $$(\sin(2x^2))'=4x\cos(2x^2).$$

Note $$\lim_{x\rightarrow0}\left(\frac{\sin(2x^2)}{2x^2}\cdot\frac{2x^2}{x^3}\right).$$

Thus, our limit it's $+\infty$ for $x\rightarrow0^+$ and it's $-\infty$ for $x\rightarrow0^-$.