Evaluate $\lim_{x \to 0^+} {\sin(1/x^2)} $ $$
 \lim_{x \to 0^+} \sin(1/x^2) = \sin(1/0) = \sin(\infty) = \infty
$$
Would this just go to infinity? Or I am missing something?

Set $t=\frac{1}{x^2}$, you get $\lim_{t \to \infty} \sin t $ which doesn't exist
