Question:
$\displaystyle\lim_{x \to0} \frac{x \tan2x - 2x \tan x}{\left(1-\cos2x\right)^{2}} $ equals :
$\displaystyle\lim_{x \to0} \frac{x \tan2x - 2x \tan x}{\left(1-\cos2x\right)^{2}} $ equals :
Updated On: Apr 2, 2026
- $\frac{1}{4}$
- 1
- $\frac{1}{2}$
- $ - \frac{1}{2}$
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The Correct Option is C
Solution and Explanation
Given:$ \displaystyle\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}} $
$\Rightarrow \displaystyle \lim _{x \rightarrow 0} \frac{\frac{2 x \tan x}{1-\tan ^{2} x}-2 x \tan x}{\left(1-1+2 \sin ^{2} x\right)^{2}} \Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{2 x \tan x}{1-\tan ^{2} x}\left(\frac{1-1+\tan ^{2} x}{4 \sin ^{4} x}\right) $
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{1}{2}\left(\frac{x}{\sin x}\right)\left(\frac{\tan ^{3} x}{x^{3}}\right)\left(\frac{x^{3}}{\sin ^{3} x}\right)=\frac{1}{2} $
$\Rightarrow \displaystyle \lim _{x \rightarrow 0} \frac{\frac{2 x \tan x}{1-\tan ^{2} x}-2 x \tan x}{\left(1-1+2 \sin ^{2} x\right)^{2}} \Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{2 x \tan x}{1-\tan ^{2} x}\left(\frac{1-1+\tan ^{2} x}{4 \sin ^{4} x}\right) $
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{1}{2}\left(\frac{x}{\sin x}\right)\left(\frac{\tan ^{3} x}{x^{3}}\right)\left(\frac{x^{3}}{\sin ^{3} x}\right)=\frac{1}{2} $
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Concepts Used:
Limits And Derivatives
Mathematically, a limit is explained as a value that a function approaches as the input, and it produces some value. Limits are essential in calculus and mathematical analysis and are used to define derivatives, integrals, and continuity.
Limits Formula:
Derivatives of a Function:
A derivative is referred to the instantaneous rate of change of a quantity with response to the other. It helps to look into the moment-by-moment nature of an amount. The derivative of a function is shown in the below-given formula.
Properties of Derivatives:
Read More: Limits and Derivatives