If f(2) = 4 and f'(2) = 1,then mathop {lim }l | vedclass

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(B) Given $f(2) = 4$ and $f'(2) = 1$.We need to evaluate $\mathop {\lim }\limits_{x 	o 2} \frac{xf(2) - 2f(x)}{x - 2}$.Adding and subtracting $2f(2)$

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$2$

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(B) Given $f(2) = 4$ and $f'(2) = 1$.We need to evaluate $\mathop {\lim }\limits_{x 	o 2} \frac{xf(2) - 2f(x)}{x - 2}$.Adding and subtracting $2f(2)$ in the numerator:$= \mathop {\lim }\limits_{x 	o 2} \frac{xf(2) - 2f(2) + 2f(2) - 2f(x)}{x - 2}$$= \mathop {\lim }\limits_{x 	o 2} \left[ \frac{f(2)(x - 2)}{x - 2} - \frac{2(f(x) - f(2))}{x - 2} ight]$$= f(2) - 2 \mathop {\lim }\limits_{x 	o 2} \frac{f(x) - f(2)}{x - 2}$$= f(2) - 2f'(2)$Substituting the values: $4 - 2(1) = 4 - 2 = 2$.Alternatively,using $L$-Hospital rule:$= \mathop {\lim }\limits_{x 	o 2} \frac{\frac{d}{dx}(xf(2) - 2f(x))}{\frac{d}{dx}(x - 2)}$$= \mathop {\lim }\limits_{x 	o 2} \frac{f(2) - 2f'(x)}{1} = f(2) - 2f'(2) = 4 - 2(1) = 2$.

If $f(2) = 4$ and $f'(2) = 1$,then $\mathop {\lim }\limits_{x \to 2} \frac{xf(2) - 2f(x)}{x - 2} = $

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