The limit of $(4-\sqrt{16-7\sin(x)})/(8x)$ at zero without using L'Hôpital

Question

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.

Answer 1

We can have $$\frac{4-\sqrt{16-7\sin x}}{8x}=\frac{(4-\sqrt{16-7\sin x})\color{red}{(4+\sqrt{16-7\sin x})}}{8x\color{red}{(4+\sqrt{16-7\sin x})}}$$ $$=\frac{7\sin x}{8x(4+\sqrt{16-7\sin x})}=\frac 78\cdot \frac{\sin x}{x}\cdot\frac{1}{4+\sqrt{16-7\sin x}}.$$ Here, note that $$\lim_{x\to 0}\frac{\sin x}{x}=1.$$

Answer 2

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.

Answer 3

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?

Answer 4

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

Answer 5

$$ \begin{aligned} \lim _{x\to 0}\left(\frac{4-\sqrt{16-7\sin \left(x\right)}}{8x}\right) & =\lim _{x\to 0}\left(\frac{4-\left(4-\frac{7x}{8}+o\left(x\right)\right)}{8x}\right) \\& = \color{red}{\frac{7}{64}} \end{aligned} $$ Solved with Taylor expansion