Question

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}$ is equal to

(a) 1

(b) $\frac{1}{2}$

(c) 0

(d) $\frac{1}{4}$

Answer 1

$\text {  } \begin{aligned}\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x} & \times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1} \\& =-\lim _{x \rightarrow 0} \frac{1+x-1}{x(\sqrt{1+x}+1)} \\& =\frac{1}{1+1}=\frac{1}{2}\end{aligned}$

So, The correct option of this question will be (B).