if f(x) has a derivative at x=a then lim x-->a [xf(a)-af(x)]/(x-a)=?

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$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) \mathrm{has} \mathrm{derivative} \mathrm{at} \mathrm{x}=\mathrm{a} \mathrm{i}.\mathrm{e}. \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{differentiable} \mathrm{at} \mathrm{x}=\mathrm{a} \\ \mathrm{We} \mathrm{have} \\ \mathrm{L}=\underset{\mathrm{x}\to \mathrm{a}}{\lim } \frac{\mathrm{xf}\left(\mathrm{a}\right)-\mathrm{af}\left(\mathrm{x}\right)}{\mathrm{x}-\mathrm{a}} \\ \mathrm{At} \mathrm{x}=\mathrm{a}, \mathrm{both} \mathrm{numerator} \mathrm{and} \mathrm{denominator}, \mathrm{become} 0, \mathrm{hence} \mathrm{it} \mathrm{is} \frac{0}{0} \mathrm{form}. \\ \mathrm{Hence} \mathrm{L}-\mathrm{Hopital} \mathrm{rule} \mathrm{can} \mathrm{be} \mathrm{applied}. \\ \mathrm{Differentiate} \mathrm{numerator} \mathrm{and} \mathrm{denominator} \mathrm{separately} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x}. \\ \mathrm{L}=\underset{\mathrm{x}\to \mathrm{a}}{\lim } \frac{\mathrm{f}\left(\mathrm{a}\right)-\mathrm{af}\text{'}\left(\mathrm{x}\right)}{1-0} \\ =\underset{\mathrm{x}\to \mathrm{a}}{\lim } \mathrm{f}\left(\mathrm{a}\right)-\mathrm{a}\text{'}\mathrm{f}\left(\mathrm{x}\right) \\ =\mathrm{f}\left(\mathrm{a}\right)-\mathrm{a}.\mathrm{f}\text{'}\left(\mathrm{a}\right) \end{gathered}$$

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