Let f: R rightarrow R be a function such that | vedclass

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(D) Let $L = \lim _{x \rightarrow 2} \frac{x^{2} f(2)-4 f(x)}{x-2}$.Since $f(2)=4$,the expression becomes $\lim _{x \rightarrow 2} \frac{4x^{2}-4 f(x)

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) Let $L = \lim _{x \rightarrow 2} \frac{x^{2} f(2)-4 f(x)}{x-2}$.Since $f(2)=4$,the expression becomes $\lim _{x \rightarrow 2} \frac{4x^{2}-4 f(x)}{x-2}$,which is of the form $\frac{0}{0}$ as $x \rightarrow 2$.Applying $L$'Hopital's Rule by differentiating the numerator and denominator with respect to $x$:$L = \lim _{x \rightarrow 2} \frac{\frac{d}{dx}(x^{2} f(2)-4 f(x))}{\frac{d}{dx}(x-2)}$$L = \lim _{x \rightarrow 2} \frac{2x f(2)-4 f^{\prime}(x)}{1}$Substitute $f(2)=4$ and $f^{\prime}(2)=1$:$L = 2(2)(4) - 4(1)$$L = 16 - 4 = 12$.

Let $f: R \rightarrow R$ be a function such that $f(2)=4$ and $f^{\prime}(2)=1$. Then,the value of $\lim _{x \rightarrow 2} \frac{x^{2} f(2)-4 f(x)}{x-2}$ is equal to:

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