Question

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

Answer 1

$\pi$

Answer 2

The limit is of the form $\frac{0}{0}$.

Use the identity $\cos^2 x = 1 - \sin^2 x$ in the argument of sine: $\pi \cos^2 x = \pi(1 - \sin^2 x) = \pi - \pi \sin^2 x$.

Use the identity $\sin(\pi - \theta) = \sin \theta$ to get $\sin(\pi \cos^2 x) = \sin(\pi - \pi \sin^2 x) = \sin(\pi \sin^2 x)$.

The limit becomes $\lim_{x\to 0} \frac{\sin(\pi \sin^2 x)}{x^2}$.

Rewrite the expression as $\frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \frac{\pi \sin^2 x}{x^2}$.

Take the limit of each part separately.

$\lim_{x\to 0} \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} = 1$ using the standard limit $\lim_{y\to 0} \frac{\sin y}{y} = 1$ with $y = \pi \sin^2 x$.

$\lim_{x\to 0} \frac{\pi \sin^2 x}{x^2} = \pi \lim_{x\to 0} \left(\frac{\sin x}{x}\right)^2 = \pi (1)^2 = \pi$ using the standard limit $\lim_{x\to 0} \frac{\sin x}{x} = 1$.

The overall limit is the product of the two limits: $1 \cdot \pi = \pi$.