Let f : R rightarrow R be a differentiable fu | vedclass

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(C) Let $L = \lim_{x 	o 2} \int_{2}^{f(x)} \frac{4t^3}{x - 2} dt$. Since $f(2) = 2$,as $x 	o 2$,the integral becomes $\int_{2}^{2} \frac{4t^3}{0} dt

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$32 f'(2)$

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(C) Let $L = \lim_{x 	o 2} \int_{2}^{f(x)} \frac{4t^3}{x - 2} dt$. Since $f(2) = 2$,as $x 	o 2$,the integral becomes $\int_{2}^{2} \frac{4t^3}{0} dt$,which is an indeterminate form of type $\frac{0}{0}$. We apply $L$'Hopital's rule and Leibniz's rule for differentiation under the integral sign: $L = \lim_{x 	o 2} \frac{\frac{d}{dx} \int_{2}^{f(x)} 4t^3 dt}{\frac{d}{dx} (x - 2)}$. Using Leibniz's rule: $\frac{d}{dx} \int_{2}^{f(x)} 4t^3 dt = 4(f(x))^3 \cdot f'(x)$. The denominator is $\frac{d}{dx} (x - 2) = 1$. Thus,$L = \lim_{x 	o 2} \frac{4(f(x))^3 \cdot f'(x)}{1}$. Substituting $x = 2$: $L = 4(f(2))^3 \cdot f'(2)$. Given $f(2) = 2$,we have $L = 4(2)^3 \cdot f'(2) = 4(8) \cdot f'(2) = 32 f'(2)$.

Let $f : R \rightarrow R$ be a differentiable function such that $f(2) = 2$. Then the value of $\lim_{x \to 2} \int_{2}^{f(x)} \frac{4t^3}{x - 2} dt$ is

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