If f: R rightarrow R is a differentiable func | vedclass

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(A) Let $L = \lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}$.Since the limit is in the $\frac{0}{0}$ form, we can apply $L$'$H$ôpital's rule or man

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$\left(1-a^2\right) f(a)$

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(A) Let $L = \lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}$.Since the limit is in the $\frac{0}{0}$ form, we can apply $L$'$H$ôpital's rule or manipulate the expression.Adding and subtracting $a f(a)$ in the numerator:$L = \lim _{x \rightarrow a} \frac{x f(a) - a f(a) + a f(a) - a f(x)}{x-a}$$L = \lim _{x \rightarrow a} \left[ \frac{f(a)(x-a)}{x-a} - a \frac{f(x)-f(a)}{x-a} \right]$$L = f(a) - a \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$By the definition of the derivative, $\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} = f^{\prime}(a)$.So, $L = f(a) - a f^{\prime}(a)$.Given $f^{\prime}(a) = a f(a)$, we substitute this into the expression:$L = f(a) - a(a f(a))$$L = f(a) - a^2 f(a)$$L = (1-a^2) f(a)$.

If $f: R \rightarrow R$ is a differentiable function at $a \in R$ such that $f^{\prime}(a)=a f(a)$, then $\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}=$

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