mathop {lim }limits_{h to 0} frac{{{{(a + h)} | vedclass

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(C) The given limit is of the form $\mathop {\lim }\limits_{h 	o 0} \frac{f(a+h) - f(a)}{h}$,which is the definition of the derivative $f'(a)$ for th

Vedclass · Accepted Answer

$2a\sin a + {a^2}\cos a$

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(C) The given limit is of the form $\mathop {\lim }\limits_{h 	o 0} \frac{f(a+h) - f(a)}{h}$,which is the definition of the derivative $f'(a)$ for the function $f(x) = x^2 \sin x$.First,find the derivative of $f(x) = x^2 \sin x$ with respect to $x$ using the product rule:$f'(x) = \frac{d}{dx}(x^2) \sin x + x^2 \frac{d}{dx}(\sin x)$$f'(x) = 2x \sin x + x^2 \cos x$Evaluating at $x = a$ gives:$f'(a) = 2a \sin a + a^2 \cos a$Alternatively,applying $L$-Hospital's rule with respect to $h$:$\mathop {\lim }\limits_{h 	o 0} \frac{\frac{d}{dh}((a+h)^2 \sin(a+h) - a^2 \sin a)}{1}$$= \mathop {\lim }\limits_{h 	o 0} (2(a+h) \sin(a+h) + (a+h)^2 \cos(a+h))$$= 2a \sin a + a^2 \cos a$

$\mathop {\lim }\limits_{h \to 0} \frac{{{{(a + h)}^2}\sin (a + h) - {a^2}\sin a}}{h} = $

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