If f(3) = 6 and f'(3) = 2,then mathop {text{L | vedclass

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(C) Given $f(3) = 6$ and $f'(3) = 2$. We need to evaluate $L = \mathop {	ext{Limit}}\limits_{x 	o 3} \frac{x f(3) - 3 f(x)}{x - 3}$. Since the form

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$0$

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(C) Given $f(3) = 6$ and $f'(3) = 2$. We need to evaluate $L = \mathop {	ext{Limit}}\limits_{x 	o 3} \frac{x f(3) - 3 f(x)}{x - 3}$. Since the form is $\frac{3 f(3) - 3 f(3)}{3 - 3} = \frac{0}{0}$,we apply $L$'$H$ôpital's rule or algebraic manipulation. Using $L$'$H$ôpital's rule: $L = \mathop {	ext{Limit}}\limits_{x 	o 3} \frac{\frac{d}{dx}(x f(3) - 3 f(x))}{\frac{d}{dx}(x - 3)}$ $L = \mathop {	ext{Limit}}\limits_{x 	o 3} \frac{f(3) - 3 f'(x)}{1}$ Substituting the values: $L = f(3) - 3 f'(3) = 6 - 3(2) = 6 - 6 = 0$. Thus,the correct option is $C$.

If $f(3) = 6$ and $f'(3) = 2$,then $\mathop {\text{Limit}}\limits_{x \to 3} \frac{x f(3) - 3 f(x)}{x - 3}$ is given by:

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