NIMCET 2017 — Mathematics PYQ

Concept:

$\cos 2x=1-2{\sin }^{2}x$

${\lim }_{x\to 0}\frac{\tan x}{x}=1$

${\lim }_{x\to 0}\frac{\sin x}{x}=1$

Calculation:

Given, lim${}_{x\to 0}\frac{x\tan x}{1-\cos x}$

$\Rightarrow$ lim${}_{x\to 0}\frac{x\tan x}{2{\sin }^2\frac{x}{2}}$

$=\frac{1}{2}$ lim${}_{x\to 0}\frac{\tan x}{x}\times \frac{\frac{x}{2}}{{\sin }^2\frac{x}{2}/{\left(\frac{x}{2}\right)}^2}{\left(\frac{x}{2}\right)}^2$

$=\frac{1}{2}$ lim${}_{x\to 0}\frac{\tan x}{x}\times \frac{4}{{\left(\frac{x}{2}\right)}^2}$

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

Hence, option (4) is correct