$\sin (x) \ e^{-x}$ limit

Question

The problem I have is:

$\lim \limits_{x \to \infty} \sin{(x)}\ e^{-x}$

Things I've tried:

1. Researching how to do this problem, I've come across kind of similar examples that use either Taylors Rule, L'Hopitals Rule, or the Squeeze Theorem. Not sure which one to use.

2. On Wolfram the anser is $0$.

3. Broke the problem apart into $\lim \limits_{x \to \infty} \sin{(x)}\lim \limits_{x \to \infty} e^{-x}$

$\lim \limits_{x \to \infty} \sin{(x)}=[-1,1]$

$\lim \limits_{x \to \infty} e^{-x}=0$

Answer 1

$|(\sin x)e^{-x}|\leq e^{-x}\rightarrow 0$ as $x\rightarrow\infty$.

Answer 2

For fun:

Let $x \gt 0:$

$0\le |\dfrac{\sin x}{e^x}| \le \dfrac{1}{e^x}\lt\dfrac{1}{x}.$

Used: $e^x = 1+x+x^2/2! +...\gt x.$

Let $\epsilon >0$ be given.

Choose $M$, real, such that $M >1/\epsilon.$

For $x\gt M$ we have:

$0\le |\dfrac{\sin x}{e^x}| \lt \dfrac{1}{x} \lt\dfrac{1}{M} \lt \epsilon.$