Question

If $f\left(x\right)=\frac{\sqrt{4+x}-2}{x}$, $x\ne 0$ be continuous at $x=0$, then $f(0)=$

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $2$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

$\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }$ $\frac{\sqrt{4+x}-2}{x}$
$=\underset{x\to 0}{\lim }$ $\left(\frac{\sqrt{4+x}-2}{x}\times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}\right)$
$=\underset{x\to 0}{\lim }$ $\frac{4+x-4}{x\left(\sqrt{4+x}+2\right)}$
$=\underset{x\to 0}{\lim }$ $\frac{x}{x\left(\sqrt{4+x}+2\right)}$
$=\underset{x\to 0}{\lim }$ $\frac{1}{\sqrt{4+x}+2}$
$=\frac{1}{2+2}$
$=\frac{1}{4}$.