ISI2015-DCG-45

Question

The value of $\underset{x \to 0}{\lim} \dfrac{\tan ^2 x – x \tan x }{\sin x}$ is

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{2}$
• C. $0$
• D. None of these

Answer 1

$\underset{x \to 0}{\lim}\dfrac{\tan^2 x-x\ \tan\ x}{\sin\ x}$

Since it is $\dfrac{0}{0}$ so we can apply L' Hospital's Rule and differentiating the numerator and denominator we get

$\underset{x \to 0}{\lim}\dfrac{2*\tan\ x* \sec^2\ x-x\ \sec^2\ x - \tan\ x}{\cos\ x}$

Applying the limit we get

$\dfrac{2*0*1-0*1 - 0}{1} =0$

$\therefore$ Option $C.$ is the correct option.

Answer 2

$\large \lim_{x->0}\frac{\tan ^{2}x-x\tan x}{sinx}$

dividing numerator and denomerator by x we get,

=$\large \lim_{x->0}\frac{\frac{\tan ^{2}x}{x}-\tan x}{\frac{\sin x}{x}}$

=$\large \frac{\lim_{x->0}\frac{\tan ^{2}x}{x}-\lim_{x->0}\tan x}{\lim_{x->0}\frac{\sin x}{x}}$

now, $\large \lim_{x->0}\frac{\sin x}{x}=1$,$\large \lim_{x->0}\tan x=0$

so we just need to evaluate ,

$\large \lim_{x->0}\frac{\tan ^{2}x}{x}$

=$\large \lim_{x->0}x.\frac{\tan ^{2}x}{x^{2}}$

=$\large \lim_{x->0}x.\lim_{x->0}\frac{\sin ^{2}x}{x^{2}}.\lim_{x->0}\frac{1}{\cos ^{2}x}$

=$0*1*1$

=$0$

so our required answer ,

=$\large \frac{0-0}{1}=0$

correct answer is option C.