The limit of $(4-\sqrt{16-7\sin(x)})/(8x)$ at zero without using L'Hôpital I stumbled across this silly limit and I am perplexed at how I can arrive to a solution by only relying on the simplest rules of limits.
$$
\lim_{x \to 0}\frac{4-\sqrt{16-7\sin(x)}}{8x}
$$
Any help is appreciated, thanks in advance.

View it as the derivative of $f(x) = \sqrt{16-7\sin x}$ at $x = 0$ or using the fact that $\sin x \approx x$ near $x = 0$, and expand the top expression binomially. 
