Limits at Infinity and Existence

Question

I am trying to prove: IF $f(−1) = 0$ and $f(1) = 2$ THEN $\lim_{x\to\infty} f(\sin(x))$ does not exist.

I considered the rule:

$$\lim_{x\to\infty} f(\sin(x)) = f(\lim_{x\to\infty} \sin(x))$$

Is it correct to use this, or is my approach all wrong?

Answer 1

Your approach is wrong.

Consider limits through the sequences $\{\frac {\pi} 2 +2n\pi\}$ and $\{-\frac {\pi} 2 +2n\pi\}$.

These sequences tend to $\infty$ and $f(\sin x))$ has different limts through these sequences.

Answer 2

It depends upon the function $f$, for example

$$f(x)=\frac x x\implies f(\sin x) \to 1$$