Question

Write the value of $\underset{x\to 0}{\lim }\frac{\sin x}{\sqrt{1+x}-1}.$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{\sin x}{\sqrt{1+x}-1}\right] \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denominator},\mathrm{we}\mathrm{get}: \\ =\underset{x\to 0}{\lim }\left[\frac{\sin x}{\left(\sqrt{1+x}-1\right)}\times \left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}\right)\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\sin x\times \left(\sqrt{1+x}+1\right)}{1+x-1}\right] \\ =\underset{x\to 0}{\lim }\left[\left(\frac{\sin x}{x}\right)\left(\sqrt{1+x}+1\right)\right] \\ =1\times \left(\sqrt{1}+1\right) \\ =2 \end{gathered}$$