If f(x) = x^2 sin frac{1}{x} for x neq 0 and | vedclass

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(A) For $x 
eq 0$,we find the derivative $f^{\prime}(x)$ using the product rule and chain rule: $f^{\prime}(x) = \frac{d}{dx} \left( x^2 \sin \frac{1

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Does not exist

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(A) For $x 
eq 0$,we find the derivative $f^{\prime}(x)$ using the product rule and chain rule: $f^{\prime}(x) = \frac{d}{dx} \left( x^2 \sin \frac{1}{x} ight) = 2x \sin \frac{1}{x} + x^2 \left( \cos \frac{1}{x} ight) \left( -\frac{1}{x^2} ight)$ $f^{\prime}(x) = 2x \sin \frac{1}{x} - \cos \frac{1}{x}$ Now,we evaluate the limit as $x ightarrow 0$: $\lim_{x ightarrow 0} f^{\prime}(x) = \lim_{x ightarrow 0} \left( 2x \sin \frac{1}{x} - \cos \frac{1}{x} ight)$ $= \lim_{x ightarrow 0} (2x \sin \frac{1}{x}) - \lim_{x ightarrow 0} \cos \frac{1}{x}$ Since $\lim_{x ightarrow 0} 2x \sin \frac{1}{x} = 0$ (by the squeeze theorem) and $\lim_{x ightarrow 0} \cos \frac{1}{x}$ does not exist because the function oscillates between $-1$ and $1$ as $x ightarrow 0$,the overall limit does not exist.

If $f(x) = x^2 \sin \frac{1}{x}$ for $x \neq 0$ and $f(0) = 0$,then find $\lim_{x \rightarrow 0} f^{\prime}(x)$.

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