LimitsmediumFree

Limits: Equal

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

Question
lim⁡x→∞x3sin⁡(1x)−2x21+3x2\displaystyle\lim_{x\to\infty}\dfrac{x^{3}\sin\left(\dfrac{1}{x}\right)-2x^{2}}{1+3x^{2}} is equal to
A00
B−13-\dfrac{1}{3}correct
C−1-1
D−23-\dfrac{2}{3}
Solution
Step 1: Divide numerator and denominator by x2x^{2}:
lim⁡x→∞xsin⁡(1/x)−21x2+3\lim_{x\to\infty}\dfrac{x\sin(1/x)-2}{\dfrac{1}{x^{2}}+3}
 Step 2: Evaluate lim⁡x→∞xsin⁡(1x)\displaystyle\lim_{x\to\infty}x\sin\left(\dfrac{1}{x}\right). Let u=1xu=\dfrac{1}{x}. As x→∞x\to\infty, u→0u\to 0:
xsin⁡(1x)=sin⁡uu→1x\sin\left(\dfrac{1}{x}\right)=\dfrac{\sin u}{u}\to 1
 Step 3: 1x2→0\dfrac{1}{x^{2}}\to 0. Step 4: Substitute:
1−20+3=−13\dfrac{1-2}{0+3}=-\dfrac{1}{3}
 Correct answer: (2) −13-\dfrac{1}{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