limx→0(1-cos2x)/(xtan2x+2x3tan3x)=

Correct answer: (B) 1/2. Given, limx→0(1-cos2x)/(xtan2x+2x3tan3x) =limx→0(2sin^2x)/xtan2x2x2x+2x/3tan3x/3x3x =limx→0(2sin^2x)/(2x^2tan2x2x)+tan3x/3x =limx→0sinxx^21/(tan2x2x+tan3x3x) =1/(1+1) =1/2

Q106 / 160 · TS EAMCET 2021 (04 Aug Shift 2) · Mathematics

$\lim_{x \to 0} \frac{(1 - \cos 2 x)}{x \tan 2 x + \frac{2 x}{3} \tan 3 x} =$

A
$- 6$
Correct answer
B
$\frac{1}{2}$
C
$0$
D
$\frac{- 6}{5}$

Solution

Given, $\lim_{x \to 0} (\frac{1 - \cos 2 x}{x \tan 2 x + \frac{2 x}{3} \tan 3 x})$

$= \lim_{x \to 0} \frac{2 \sin^{2} x}{x \frac{\tan 2 x}{2 x} 2 x + \frac{2 x}{3} \frac{\tan 3 x}{3 x} 3 x}$

$= \lim_{x \to 0} \frac{2 \sin^{2} x}{2 x^{2} (\frac{\tan 2 x}{2 x} + \frac{\tan 3 x}{3 x})}$

$= \lim_{x \to 0} {(\frac{\sin x}{x})}^{2} \frac{1}{(\frac{\tan 2 x}{2 x} + \frac{\tan 3 x}{3 x})}$

$= \frac{1}{1 + 1}$

$= \frac{1}{2}$

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

limx→01-cos2xxtan2x+2x3tan3x=

Solution

Relevant Questions

$\lim_{x \to 0} \frac{(1 - \cos 2 x)}{x \tan 2 x + \frac{2 x}{3} \tan 3 x} =$