If f: R rightarrow R is such that f(3)=16 and | vedclass

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(A) Given: $f(3)=16$ and $f^{\prime}(3)=4$. We need to evaluate the limit: $L = \lim _{x \rightarrow 3} \frac{x f(3)-3 f(x)}{x-3}$. Substituting $x=3$

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$4$

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(A) Given: $f(3)=16$ and $f^{\prime}(3)=4$. We need to evaluate the limit: $L = \lim _{x \rightarrow 3} \frac{x f(3)-3 f(x)}{x-3}$. Substituting $x=3$,we get the form $\frac{3f(3)-3f(3)}{3-3} = \frac{0}{0}$. Applying $L^{\prime}$Hospital's Rule by differentiating the numerator and denominator with respect to $x$: $L = \lim _{x \rightarrow 3} \frac{\frac{d}{dx}(x f(3)-3 f(x))}{\frac{d}{dx}(x-3)}$. $L = \lim _{x \rightarrow 3} \frac{f(3)-3 f^{\prime}(x)}{1}$. Substituting $x=3$: $L = f(3)-3 f^{\prime}(3) = 16 - 3(4) = 16 - 12 = 4$.

If $f: R \rightarrow R$ is such that $f(3)=16$ and $f^{\prime}(3)=4$,then find the value of $\lim _{x \rightarrow 3} \frac{x f(3)-3 f(x)}{x-3}$.

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