LimitseasyFree

Limits — JEE Maths practice question

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

Question
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
A−148-\dfrac{1}{48}correct
B−43-\dfrac{4}{3}
C−163-\dfrac{16}{3}
D112\dfrac{1}{12}
Solution
Step 1: Combine the fractions over a common denominator:
lim⁡h→01h[18+h3−12]=lim⁡h→02−8+h32h8+h3\lim_{h\to 0}\dfrac{1}{h}\left[\dfrac{1}{\sqrt[3]{8+h}}-\dfrac{1}{2}\right]=\lim_{h\to 0}\dfrac{2-\sqrt[3]{8+h}}{2h\sqrt[3]{8+h}}
 Step 2: Rationalise the numerator using the identity a−b=a3−b3a2+ab+b2a-b=\dfrac{a^{3}-b^{3}}{a^{2}+ab+b^{2}} with a=2,  b=8+h3a=2,\;b=\sqrt[3]{8+h}:
2−8+h3=8−(8+h)4+28+h3+(8+h)23=−h4+28+h3+(8+h)232-\sqrt[3]{8+h}=\dfrac{8-(8+h)}{4+2\sqrt[3]{8+h}+\sqrt[3]{(8+h)^{2}}}=\dfrac{-h}{4+2\sqrt[3]{8+h}+\sqrt[3]{(8+h)^{2}}}
 Step 3: Substitute back:
lim⁡h→0−h2h8+h3⋅[4+28+h3+(8+h)23]\lim_{h\to 0}\dfrac{-h}{2h\sqrt[3]{8+h}\cdot\left[4+2\sqrt[3]{8+h}+\sqrt[3]{(8+h)^{2}}\right]}
 Step 4: Cancel hh and substitute h=0h=0:
=−12⋅2⋅(4+4+4)=−14⋅12=−148=\dfrac{-1}{2\cdot 2\cdot(4+4+4)}=\dfrac{-1}{4\cdot 12}=-\dfrac{1}{48}
 Correct answer: (1) −148-\dfrac{1}{48}
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Limits · easy
x x ( x+1 - x ) equal to
Limits · easy
x r=1 10 (x+r) 2010 (x 1006 +1)(2x 1004 +1) , is equal to
Limits · easy
x 0 ae x +b x+ce -x e 2x -2e x +1 =4 . Then
Limits · easy
The value of x 1 x 1/3 -x 1/4 x 3 -1 is

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