Discussion
\(\displaystyle\lim_{x \to 0} \frac{\sin(\pi \sin^{2}x)}{x^{2}}\) =
Answer: B
The limit is \(\pi\).
As \(x \to 0\), \(\pi\sin^{2}x \to 0\), so \(\sin(\pi\sin^{2}x) \approx \pi\sin^{2}x\):
\[ \begin{aligned} \lim_{x \to 0} \frac{\sin(\pi\sin^{2}x)}{\pi\sin^{2}x} \cdot \frac{\pi\sin^{2}x}{x^{2}} &= 1 \cdot \pi \cdot 1 \\ &= \pi \end{aligned} \]
\(\pi^{2}\) and \(\frac{\pi}{2}\) come from mishandling the factor \(\pi\) or the square.
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