Using epsilon-delta definition to find limit involving $\sin x$

Question

I need to show $$\lim_{x \to \infty}\frac{x^2-16}{x+\sin x}=0.$$ I am mostly unsure how to simplify the expression $$\lim_{x \to \infty}\left|\frac{x^2-16}{x+\sin x}\right|<\epsilon$$ due to the $\sin x$ involved. I got up to $$\left|\frac{x^2-16}{x+\sin x}\right|<\left|\frac{x^2-16}{x+1}\right|,$$ but I'm not sure whether this is the right or how to proceed.

Answer 1

The limit is $\infty$, not $0$.

Recall that $0\le x+\sin(x)\le x+x=2x$. Then, given any number $B>0$, however large, and for $x\ge 4\sqrt{2}$

$$\frac{x^2-16}{x+\sin(x)}\ge \frac{x^2-16}{2x}\ge \frac14 x>B$$

whenever $|x|>\max(4\sqrt{2},4B)$. And we are done!