LimitseasyFree

Limits: 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
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Limits · easy
lim⁡h→0[1h8+h3−12h]\displaystyle\lim_{h\to 0}\left[\dfrac{1}{h\sqrt[3]{8+h}}-\dfrac{1}{2h}\right] is equal to
Limits · easy
lim⁡x→∞∑r=110(x+r)2010(x1006+1)(2x1004+1) \displaystyle\lim_{x\to\infty}\dfrac{\sum_{r=1}^{10}(x+r)^{2010}}{(x^{1006}+1)(2x^{1004}+1)}\, is equal to
Limits · easy
The value of lim⁡x→1x1/3−x1/4x3−1\displaystyle \lim_{x\to 1}\dfrac{x^{1/3}-x^{1/4}}{x^{3}-1} is
Limits · easy
lim⁡x→0ex−esin⁡x2(x−sin⁡x)= ?\displaystyle\lim_{x\to 0}\dfrac{e^{x}-e^{\sin x}}{2(x-\sin x)}=\,?

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath