If f(5)=7 and f'(5)=7, then lim_{x rightarrow | vedclass

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(D) Let $L = \lim_{x \rightarrow 5} \frac{x f(5)-5 f(x)}{x-5}$.Since the limit is in the $\frac{0}{0}$ form, we apply $L$'$H$ôpital's rule by differen

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-$28$

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(D) Let $L = \lim_{x \rightarrow 5} \frac{x f(5)-5 f(x)}{x-5}$.Since the limit is in the $\frac{0}{0}$ form, we apply $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:$L = \lim_{x \rightarrow 5} \frac{\frac{d}{dx}(x f(5)-5 f(x))}{\frac{d}{dx}(x-5)}$$L = \lim_{x \rightarrow 5} \frac{f(5)-5 f'(x)}{1}$Substituting $x=5$:$L = f(5)-5 f'(5)$Given $f(5)=7$ and $f'(5)=7$:$L = 7 - 5(7) = 7 - 35 = -28$.

If $f(5)=7$ and $f'(5)=7$, then $\lim_{x \rightarrow 5} \frac{x f(5)-5 f(x)}{x-5}$ is given by

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