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

JEE MATHEMATICS MCQ: LIMITS

1 Easy

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Application +4 | -1

Let

f\,:\,R \to R

be a function such that

f\left( 2 \right) = 4

and

f'\left( 2 \right) = 1

. Then, the value of

\lim\limits_{x \to 2} \frac{{{x^2}f\left( 2 \right) - 4f\left( x \right)}}{{x - 2}}

is equal to

View answer and explanation

Correct option: D

Option D (12) is correct because applying L'Hopital's rule to the limit and using the given values of f(2) and f'(2), we can find the value of the limit.

JEE MATHEMATICS MCQ: LIMITS

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