Let f(2) = 4 and f'(2) = 4,then mathop {lim } | vedclass

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(C) Let $L = \mathop {\lim }\limits_{x 	o 2} \frac{{xf(2) - 2f(x)}}{{x - 2}}$.Since the limit is of the form $\frac{0}{0}$ at $x = 2$,we can apply $L

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$-4$

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(C) Let $L = \mathop {\lim }\limits_{x 	o 2} \frac{{xf(2) - 2f(x)}}{{x - 2}}$.Since the limit is of the form $\frac{0}{0}$ at $x = 2$,we can apply $L$'$H$ôpital's rule or manipulate the expression.Adding and subtracting $2f(2)$ in the numerator:$L = \mathop {\lim }\limits_{x 	o 2} \frac{{xf(2) - 2f(2) + 2f(2) - 2f(x)}}{{x - 2}}$$L = \mathop {\lim }\limits_{x 	o 2} \left[ \frac{f(2)(x - 2)}{x - 2} - 2\frac{f(x) - f(2)}{x - 2} ight]$$L = f(2) - 2 \mathop {\lim }\limits_{x 	o 2} \frac{f(x) - f(2)}{x - 2}$$L = f(2) - 2f'(2)$Given $f(2) = 4$ and $f'(2) = 4$:$L = 4 - 2(4) = 4 - 8 = -4$.

Let $f(2) = 4$ and $f'(2) = 4$,then $\mathop {\lim }\limits_{x \to 2} \,\frac{{xf(2) - 2f(x)}}{{x - 2}}$ equals

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