Limits
Trigonometric Limits
JEE Main 2014
Grade 11

Question:

$\lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2}$
(1) $\pi$
(2) $-\pi$
(3) $\frac{\pi}{2}$
(4) $1$

Step-by-Step Solution

Key Concept: Use $\cos^2 x = 1 - \sin^2 x$, so $\sin(\pi \cos^2 x) = \sin(\pi - \pi \sin^2 x) = \sin(\pi \sin^2 x)$. Then $\frac{\sin(\pi \sin^2 x)}{x^2} = \pi \cdot \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \frac{\sin^2 x}{x^2} \to \pi \cdot 1 \cdot 1 = \pi$.
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Correct Answer: (1)

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