Question

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

Answer 1

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

$\therefore f\left(0\right)$

$=\underset{x\to 0}{\lim }f\left(x\right)$

$=\underset{x\to 0}{\lim }\frac{2-\sqrt{x+4}}{\sin 2x}$

$=\underset{x\to 0}{\lim }\frac{2-\sqrt{x+4}}{\sin 2x}\times \frac{2+\sqrt{x+4}}{2+\sqrt{x+4}}$

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

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

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

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

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

$=-\frac{1}{2}\times \frac{1}{4}$

$=-\frac{1}{8}$

Thus, the value of f(0) is $-\frac{1}{8}$.


If $f\left(x\right)=\frac{2-\sqrt{x+4}}{\sin 2x},x\ne 0,$ is continuous at x = 0, then f(0) = $\underline{-\frac{1}{8}}$.