lim _{x rightarrow 0} x^2 sin left(frac{pi}{x | vedclass

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(B) We use the Squeeze Theorem to evaluate the limit.We know that for all $x 
eq 0$,$-1 \leq \sin \left(\frac{\pi}{x}ight) \leq 1$.Multiplying the

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$0$

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(B) We use the Squeeze Theorem to evaluate the limit.We know that for all $x 
eq 0$,$-1 \leq \sin \left(\frac{\pi}{x}ight) \leq 1$.Multiplying the inequality by $x^2$ (since $x^2 > 0$ for $x 
eq 0$),we get:$-x^2 \leq x^2 \sin \left(\frac{\pi}{x}ight) \leq x^2$.Now,taking the limit as $x ightarrow 0$ on all sides:$\lim _{x ightarrow 0} (-x^2) \leq \lim _{x ightarrow 0} x^2 \sin \left(\frac{\pi}{x}ight) \leq \lim _{x ightarrow 0} x^2$.Since $\lim _{x ightarrow 0} (-x^2) = 0$ and $\lim _{x ightarrow 0} x^2 = 0$,by the Squeeze Theorem:$\lim _{x ightarrow 0} x^2 \sin \left(\frac{\pi}{x}ight) = 0$.

$\lim _{x \rightarrow 0} x^2 \sin \left(\frac{\pi}{x}\right)$ is equal to

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