<s>[INST] 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.
 [/INST] Multiply the numerator and denominator by $4+\sqrt{16-7\sin x}$ to get 
\begin{align}
& \lim_{x \to 0}\dfrac{4-\sqrt{16-7\sin x}}{8x} \\[8pt]
= {} &\lim_{x \to 0}\dfrac{4-\sqrt{16-7\sin x}}{8x} \cdot \dfrac{4+\sqrt{16-7\sin x}}{4+\sqrt{16-7\sin x}} \\[8pt]
= {} & \lim_{x \to 0}\dfrac{4^2 - (16-7\sin x)}{8x(4+\sqrt{16-7\sin x})} \\[8pt]
= {} & \lim_{x \to 0}\dfrac{7\sin x}{8x(4+\sqrt{16-7\sin x})} \\[8pt]
= {} & \lim_{x \to 0}\dfrac{\sin x}{x} \cdot \lim_{x \to 0}\dfrac{7}{8(4+\sqrt{16-7\sin x})}
\end{align}
Can you take it from here?
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