lim_h → 0 (f(2h + 2 + h^2) - f(2))/(f(h - h^2 + 1) - f(1)), given that f'(2) = 6 and f'(1) = 4.

lim_h → 0 (f(2h + 2 + h^2) - f(2))/(f(h - h^2 + 1) - f(1)),… — JEE Main 2003 Mathematics PYQ

AIEEE 2003

\lim_{h \to 0} \frac{f(2h + 2 + h^2) - f(2)}{f(h - h^2 + 1) - f(1)}

, given that

f'(2) = 6

and

f'(1) = 4

A.
does not exist
B.
is equal to $-3/2$
C.
is equal to $3/2$
D.
is equal to $3$

Show answer and solution

Correct option: D

Solution

Apply L'Hôpital's rule (or equivalently, use the definition of derivative). Both numerator and denominator → 0 as h → 0.

Differentiate numerator w.r.t. h: f'(2h + 2 + h²)·(2 + 2h) → f'(2)·2 = 6·2 = 12.

Differentiate denominator w.r.t. h: f'(h - h² + 1)·(1 - 2h) → f'(1)·1 = 4·1 = 4.

Limit = 12/4 = 3.

[AIEEE 2003]

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