Let f(x) be a differentiable function at x=a | vedclass

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(B) Given $f^{\prime}(a) = 2$ and $f(a) = 4$. We need to evaluate $L = \lim_{x \rightarrow a} \frac{x f(a) - a f(x)}{x - a}$. Since the limit is of th

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$4 - 2a$

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(B) Given $f^{\prime}(a) = 2$ and $f(a) = 4$. We need to evaluate $L = \lim_{x \rightarrow a} \frac{x f(a) - a f(x)}{x - a}$. Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$: $L = \lim_{x \rightarrow a} \frac{\frac{d}{dx}(x f(a) - a f(x))}{\frac{d}{dx}(x - a)}$ $L = \lim_{x \rightarrow a} \frac{f(a) - a f^{\prime}(x)}{1}$ Substituting $x = a$: $L = f(a) - a f^{\prime}(a)$ $L = 4 - a(2) = 4 - 2a$.

Let $f(x)$ be a differentiable function at $x=a$ with $f^{\prime}(a)=2$ and $f(a)=4$. Then $\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}$ equals ...... .

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