I'm having difficulty writing an $\epsilon - \delta$ proof for the following limit:
$\lim_{x\to 4} \frac{x^2-16}{x+\sin x} = 0$
I've factored it to $\frac{(x+4)(x-4)}{x+\sin x} = 0$
and guessed that I need $\delta = \frac{2}{5}\epsilon$ for $|x-4| < \delta \implies |\frac{x^2-16}{x+\sin x}| < \epsilon$
I've also bounded $|x+4|$ by $\delta + 8$ but I don't know how to control $|x + \sin x|$.
$$|x-4| < \delta$$
$$4-\delta < x < 4+ \delta$$
$$3 - \delta< x+ \sin x < 5 + \delta$$
$$\frac{1}{5+\delta} < \frac{1}{x+\sin x} < \frac{1}{3-\delta}$$
If $\delta < 1$, $-\delta > -1$, $3-\delta > $2, $\frac{1}{3-\delta} < \frac12$
$$\left| \frac{x^2-16}{x+\sin x}\right| \leq \frac12 |x^2-16|$$
Also, if $\delta < 1$, $|x+4|<9,$
$$\frac12 |x^2-16|\leq \frac 92 |x-4|$$
Hence, for example, I can choose
$$\delta = \min ( \frac12, \frac29 \epsilon)$$
