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.

$$ \lim_{x\rightarrow 0}\frac{4-\sqrt{16-7\sin x}}{8x}=\\
\lim_{x\rightarrow 0}\frac{4-\sqrt{16-7\sin x}}{8x}\frac{4+\sqrt{16-7\sin x}}{4+\sqrt{16-7\sin x}}=\\
\lim_{x\rightarrow 0}\frac{16-{16+7\sin x}}{8x}\frac{1}{4+\sqrt{16-7\sin x}}=\\
\lim_{x\rightarrow 0}\frac{{7\sin x}}{8x}\frac{1}{4+\sqrt{16-7\sin x}}=\\
\frac{7}{8}\lim_{x\rightarrow 0}\frac{{\sin x}}{x}\frac{1}{4+\sqrt{16-7\sin x}}=\\(\lim_{x\rightarrow 0}\frac{{\sin x}}{x}=1)\\
=\frac{7}{8}\lim_{x\rightarrow 0}\frac{1}{4+\sqrt{16-7\sin x}}=\\
\frac{7}{8}\lim_{x\rightarrow 0}\frac{1}{4+\sqrt{16-7(0)}}=\\
\frac{7}{8}\frac{1}{8}=\\
\frac{7}{64}
$$
