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Limits — JEE Maths practice question

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}
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