Let $f''\left( x \right)$ be continuous at $x = 0$ and $f''\left( 0 \right) = 4$. Then $\mathop {\lim }\limits_{x \to 0} \frac{{2f\left( x \right) - 3f\left( {2x} \right) + f\left( {4x} \right)}}{{{x^2}}}$ is equal to

JEE MATHEMATICS MCQ: LIMITS

1 Easy

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Analyzing +4 | -1

Let

f''\left( x \right)

be continuous at

x = 0

and

f''\left( 0 \right) = 4

. Then

\lim\limits_{x \to 0} \frac{{2f\left( x \right) - 3f\left( {2x} \right) + f\left( {4x} \right)}}{{{x^2}}}

is equal to

View answer and explanation

Correct option: C

Option C (12) is correct because applying L'Hopital's rule twice and using the given condition

f''(0) = 4

results in the limit being equal to 12.

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