Question

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

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

(b) 2

(c) 0

(d) 1

Answer 1

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

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\sqrt{1+x}-1}{x} \\ =\underset{x\to 0}{\lim }\frac{\left(\sqrt{1+x}-1\right)\left(\sqrt{1+x}+1\right)}{\left(\sqrt{1+x}+1\right)x} \\ =\underset{x\to 0}{\lim }\frac{1+x-1}{x\sqrt{1+x+1}} \\ =\underset{x\to 0}{\lim }\frac{1}{x\left(\sqrt{1+x}+1\right)} \\ =\frac{1}{2} \end{gathered}$$