Question

$\underset{x\to 0}{\lim }\frac{\sin x}{\sqrt{1+x}-\sqrt{1-x}}\mathrm{is}$

​(a) 2
(b) 0
(c) 1
(d) –1

Answer 1

$$\begin{gathered} \begin{array}{c}\lim \\ x\to 0\end{array}\frac{\sin x}{\sqrt{1+x}-\sqrt{1-x}} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{\sin x}{\sqrt{1+x}-\sqrt{1-x}}\times \frac{\left(\sqrt{1+x}+\sqrt{1-x}\right)}{\left(\sqrt{1+x}+\sqrt{1-x}\right)} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{\sin x}{{\left(\sqrt{1+x}\right)}^2-{\left(\sqrt{1-x}\right)}^2}\times \sqrt{1+x}+\sqrt{1-x} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{\sin x}{1+x-1+x}\times \sqrt{1+x}+\sqrt{1-x} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{\sin x}{2x}\times \sqrt{1+x}+\sqrt{1-x} \\ \mathrm{i}.\mathrm{e}.\frac{1}{2}\left(\begin{array}{c}\lim \\ x\to 0\end{array}\frac{\sin x}{x}\right)\left(\begin{array}{c}\lim \\ x\to 0\end{array}\sqrt{1+x}+\sqrt{1-x}\right) \\ \mathrm{i}.\mathrm{e}.\frac{1}{2}\times 1\times \left(\sqrt{1}+\sqrt{1}\right) \\ \mathrm{i}.\mathrm{e}.\frac{1}{2}\times 2=1 \end{gathered}$$

Hence, the correct answer is option C.