Let f:R→R be a function such that f(2)=4 and f1(2)=1.Then the value of limx→2x2f(2)−4f(x)x−2 is equal to

limx→2x2f(2)−4f(x)x−2 [00from]Apply L Hopital Rulelimx→2(2xf(2)−4f'(x)1)=4(4)−41=12

Let f:R→R be a function such that f(2)=4 and f1(2)=1.Then the value of limx→2x2f(2)−4f(x)x−2 is equal to

limx→2x2f(2)−4f(x)x−2 [00from]Apply L Hopital Rulelimx→2(2xf(2)−4f'(x)1)=4(4)−41=12

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Let $f : R \to R$ be a function such that $f (2) = 4$ and $f^{1} (2) = 1$ .Then the value of $\lim_{x \to 2} \frac{x^{2} f (2) - 4 f (x)}{x - 2}$ is equal to

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$\lim_{x \to 2} \frac{x^{2} f (2) - 4 f (x)}{x - 2}$ $[\frac{0}{0} f r o m]$
Apply L Hopital Rule
$\lim_{x \to 2} (\frac{2 x f (2) - 4 f' (x)}{1}) = \frac{4 (4) - 4}{1} = 12$

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Let f:R→R be a function such that f(2)=4 and f1(2)=1.Then the value of limx→2x2f(2)−4f(x)x−2 is equal to

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Let $f : R \to R$ be a function such that $f (2) = 4$ and $f^{1} (2) = 1$ .Then the value of $\lim_{x \to 2} \frac{x^{2} f (2) - 4 f (x)}{x - 2}$ is equal to