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)

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