$f(x) = \begin{cases} \frac{\sin\left(\pi\cos^2\left(\tan(\sin x)\right)\right)}{\pi x^2}, & x \neq 0 \\ k, & x = 0 \end{cases}$
The value of $k$ such that $f$ is continuous at $x = 0$, is
Step-by-Step Solution
Key Concept: Recognize trigonometric identities and use Taylor series approximations near $x = 0$ to evaluate limits.
Evaluate $k = \lim_{x \to 0} \frac{\sin(\pi\cos^2(\tan(\sin x)))}{\pi x^2}$. Using the identity $\sin(\pi - \theta) = \sin\theta$, rewrite the numerator as $\sin(\pi\sin^2(\tan(\sin x)))$. As $x \to 0$, $\sin x \sim x$, $\tan x \sim x$, so $\sin^2(\tan(\sin x)) \sim x^2$. Therefore $\sin(\pi x^2) \sim \pi x^2$, giving $k = \frac{\pi x^2}{\pi x^2} = 1$.
Correct Answer: [A-p, q] [B-p, s] [C-q] [D-r, t]