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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If f′(2)=6f'(2) = 6 and f′(1)=4f'(1) = 4, then lim⁡h→0f(2h+2+h2)−f(2)f(h−h2+1)−f(1)\displaystyle\lim_{h \to 0} \frac{f(2h+2+h^2)-f(2)}{f(h-h^2+1)-f(1)} is.
Solution
Answer: 3
Step 1: Identify the indeterminate form As h→0h \to 0, both numerator and denominator tend to 00. Differentiating with respect to hh:
L=lim⁡h→0f′(2h+2+h2)⋅(2+2h)f′(h−h2+1)⋅(1−2h)L = \lim_{h \to 0}\frac{f'(2h+2+h^2)\cdot(2+2h)}{f'(h-h^2+1)\cdot(1-2h)}
 Step 2: Evaluate at h=0h = 0
L=f′(2)⋅2f′(1)⋅1=6×24×1=3L = \frac{f'(2) \cdot 2}{f'(1) \cdot 1} = \frac{6 \times 2}{4 \times 1} = 3
 Answer: 3
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