Question

The value of $f\left(0\right)$ for the function $f\left(x\right)=\frac{2-\sqrt{(x+4)}}{\sin 2x},x\ne 0$ is continuous at $x=0$ is

• A. $\frac{1}{8}$
• B. $\frac{1}{4}$
• C. $\frac{-1}{8}$
• D. $-\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{-1}{8}$

Since, on applying the limits, the function is of $\frac{0}{0}$ form,
We apply L-Hospital's Rule,

$f\left(0\right)={lim}_{x\to 0}\frac{0-\frac{1}{2{(x+4)}^{0.5}}}{2cos2x}$

Now applying the limit, we get

$f\left(0\right)=\frac{\frac{-1}{2\times 2}}{2}=\frac{-1}{8}$