Question

$\displaystyle \lim_{x\to0}\left(\frac{\tan \,x}{\sqrt{2x +4} - 2}\right)$ is equal to

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

The Correct Option is A

$\displaystyle\lim_{x \rightarrow 0}\left(\frac{\tan \,x}{\sqrt{2 x+4}-2}\right) \,\,\,\left[\frac{0}{0}\right]$ form
Applying L' Hopitals' rule, we get
$\displaystyle\lim _{x \rightarrow 0}\left(\frac{\sec ^{2} x}{\frac{2}{2x\sqrt{2x +4}}-0}\right)$
$=\displaystyle\lim _{x \rightarrow 0}\left((\sqrt{2 x+4}) \sec ^{2} x\right)$
$=\sqrt{2 \times 0+4} \times 1=2$