Q107 / 180 · KCET 2020 · Mathematics

$\lim _{x \rightarrow 0}\left(\frac{\tan x}{\sqrt{2 x+4}-2}\right)$ is equal to

  • Correct answer
    A
    2
  • B
    3
  • C
    4
  • D
    6

Solution

We have, $\lim _{x \rightarrow 0}\left(\frac{\tan x}{\sqrt{2 x+4}-2}\right)$ $=\lim _{x \rightarrow 0} \frac{(\tan x)(\sqrt{2 x+4}+2)}{(2 x+4)-4}$ $=\lim _{x \rightarrow 0} \frac{\tan x(\sqrt{2 x+4}+2)}{2 x}$ $=\frac{1}{2} \times(\sqrt{4}+2)=\frac{1}{2}(2+2)=2$

Hence, the correct answer is option: 'A'.

Relevant Questions