Given $f(x) = e^{-x} \sin x, $ Find $\lim\limits_{x \rightarrow \infty} f(x)$ if it exists. Justify using limits definition. 
Given $$f(x) = e^{-x} \sin x, $$
Find $\lim\limits_{x \rightarrow \infty} f(x)$ if it exists. Justify using directly the following definition:
$\lim\limits_{x \rightarrow \infty} f(x)=L$ if $f$ is defined on an interval $(a, \infty)$ and for each $\epsilon >0$ there is a number $\beta$ such that:
$$| f(x) - L| < \epsilon \quad\text{if}\quad x> \beta.$$

Taking the limit:
$$ \lim\limits_{x \rightarrow \infty} e^{-x} \sin x = 0$$
By definition for any $\epsilon > 0$ we have:
$$ |e^{-x} \sin x - 0|< \epsilon$$
$$ |e^{-x} \sin x |< \epsilon$$
Let's find some $M>0$ in terms of $x$ such that:
$$|e^{-x} \sin x | \leq M < \epsilon$$
As $|\sin x |$ fluctuates from $0$ to $1$, $$|e^{-x} \sin x | \leq |e^{-x}|$$
Since $e^{-x}>0 ,\forall x$, we have
$$|e^{-x} \sin x | \leq e^{-x}$$
Let's solve for $x$
$$e^{-x} < \epsilon$$
$$\ln (e^{-x}) < \ln (\epsilon)$$
$$-x< \ln \epsilon$$
$$x> -\ln \epsilon$$
It follows that for any arbitrary $\epsilon>0$, the limit $|f(x) - 0|< \epsilon$ is true when $x>-\ln \epsilon$
Is this a correct reasoning? can it be improved?
Any input is much appreciated

Fix $\varepsilon>0$. Then
$$
\left|\frac{\sin x}{e^x}-0\right| \le \frac{1}{e^x} \le \varepsilon
$$
whenever $x \ge -\log \varepsilon$. (In particular, yes the reasoning is correct.)
