Let f : R \[\to\] R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of \[\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}\] is equal to :
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Official solution
This limit can be solved using L'Hopital's Rule, which states that for the limit of the form 0/0 or \pm \infty /\pm \infty , the limit can be found by taking the derivative of the numerator and the derivative of the denominator separately.
\[\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}\] is in the indeterminate form, and we are given that f(2) = 4 and f'(2) = 1, so we can apply L'Hopital's rule.
Taking the derivative of the numerator and the denominator, we get :
Numerator: derivative of x2f(2)−4f(x) is 2xf(2)−4f′(x).
Denominator: derivative of x−2 is 1.
So, the limit becomes :
\[\mathop {\lim }\limits_{x \to 2} {{{2xf(2) - 4f'(x)}} \over 1} = 2 \times 2 \times f(2) - 4 \times f'(2) = 16 - 4 = 12.\]
Therefore, Option D, 12, is the correct answer.