If f(5)=7 and f'(5)=7, then limx→5 xf(5)−5f(x)x−5 is given by

limx→5 xf(5)−5f(x)x−500 form =limx→5 f(5)−5f′(x)1 [Using L' Hospital's Rule]=f(5)−5f′(5)=7−5×7=−28

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If $f (5) = 7$ and $f' (5) = 7$ , then $lim_{x \to 5} \frac{x f (5) - 5 f (x)}{x - 5}$ is given by

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answer is -28.

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Detailed Solution

$lim_{x \to 5} \frac{x f (5) - 5 f (x)}{x - 5} (\frac{0}{0} form)$

$= lim_{x \to 5} \frac{f (5) - 5 f^{'} (x)}{1}$ [Using L' Hospital's Rule]

$= f (5) - 5 f^{'} (5) = 7 - 5 \times 7 = - 28$

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