Question:medium
$\displaystyle\lim_{x \to 0}\frac{1 - \cos 4x}{\tan^2 2x}$ is equal to
$\displaystyle\lim_{x \to 0}\frac{1 - \cos 4x}{\tan^2 2x}$ is equal to
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A quick method: Use $1-\cos\theta \approx \dfrac{\theta^2}{2}$ and $\tan\theta \approx \theta$ near $\theta = 0$. So $\dfrac{1-\cos 4x}{\tan^2 2x} \approx \dfrac{(4x)^2/2}{(2x)^2} = \dfrac{8x^2}{4x^2} = 2$.
Updated On: Apr 24, 2026
- 3
- $\dfrac{1}{4}$
- $\dfrac{1}{2}$
- 2
- 0
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The Correct Option is D
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