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Limits — JEE Maths practice question

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

Question
lim⁡x→05sin⁡x−7sin⁡2x+3sin⁡3xx2sin⁡x\displaystyle\lim_{x\to 0}\dfrac{5\sin x-7\sin 2x+3\sin 3x}{x^{2}\sin x} is equal to
A−5-5correct
B55
C00
D∞\infty
Solution
Step 1: Use Taylor series: sin⁡u=u−u36+…\sin u=u-\dfrac{u^{3}}{6}+\ldots. So:
5sin⁡x=5(x−x36+…)=5x−5x36+…5\sin x=5\left(x-\dfrac{x^{3}}{6}+\ldots\right)=5x-\dfrac{5x^{3}}{6}+\ldots
7sin⁡2x=7(2x−8x36+…)=14x−56x36+…7\sin 2x=7\left(2x-\dfrac{8x^{3}}{6}+\ldots\right)=14x-\dfrac{56x^{3}}{6}+\ldots
3sin⁡3x=3(3x−27x36+…)=9x−81x36+…3\sin 3x=3\left(3x-\dfrac{27x^{3}}{6}+\ldots\right)=9x-\dfrac{81x^{3}}{6}+\ldots
 Step 2: Combine:
5sin⁡x−7sin⁡2x+3sin⁡3x=(5−14+9)x+(−56+566−816)x3+…5\sin x-7\sin 2x+3\sin 3x=(5-14+9)x+\left(-\dfrac{5}{6}+\dfrac{56}{6}-\dfrac{81}{6}\right)x^{3}+\ldots
=0⋅x+−5+56−816x3+…=−306x3+…=−5x3+…=0\cdot x+\dfrac{-5+56-81}{6}x^{3}+\ldots=\dfrac{-30}{6}x^{3}+\ldots=-5x^{3}+\ldots
 Step 3: Divide by x2sin⁡xx^{2}\sin x. Since sin⁡x∼x\sin x\sim x, denominator ∼x3\sim x^{3}:
lim⁡x→0−5x3+…x2⋅x=−5x3x3=−5\lim_{x\to 0}\dfrac{-5x^{3}+\ldots}{x^{2}\cdot x}=\dfrac{-5x^{3}}{x^{3}}=-5
 Correct answer: (1) −5-5
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