Limits and Derivatives: PYQ Walkthrough (6)

hard • 3 min read
Tags Limits And Derivatives

Question

(JEE Main 2023 Shift 2) Evaluate: lim⁡x→0sin⁡(3x)−3sin⁡xx3\displaystyle \lim_{x \to 0} \frac{\sin(3x) - 3\sin x}{x^3}.

Solution — Step by Step

At x=0x = 0: numerator =sin⁡0−3sin⁡0=0= \sin 0 - 3\sin 0 = 0. Denominator =0= 0. So we have 0/00/0 form — L’Hôpital or Taylor series both work.

Recall sin⁡x=x−x3/6+x5/120−…\sin x = x - x^3/6 + x^5/120 - \ldots

sin⁡(3x)=3x−(3x)3/6+…=3x−9x3/2+…\sin(3x) = 3x - (3x)^3/6 + \ldots = 3x - 9x^3/2 + \ldots

3sin⁡x=3x−3x3/6+…=3x−x3/2+…3\sin x = 3x - 3x^3/6 + \ldots = 3x - x^3/2 + \ldots

sin⁡(3x)−3sin⁡x=(3x−9x3/2)−(3x−x3/2)+O(x5)\sin(3x) - 3\sin x = (3x - 9x^3/2) - (3x - x^3/2) + O(x^5)

=−9x3/2+x3/2=−8x3/2=−4x3= -9x^3/2 + x^3/2 = -8x^3/2 = -4x^3 (to leading order)

lim⁡x→0−4x3+O(x5)x3=−4\lim_{x \to 0} \frac{-4x^3 + O(x^5)}{x^3} = -4

Limit =−4= -4.

Why This Works

Taylor series turns trigonometric limits into polynomial arithmetic. The first non-zero term in the numerator’s expansion determines the limit when divided by the right power of xx.

For 0/00/0 forms involving trig, exponential, and logarithmic functions, Taylor is faster than L’Hôpital — especially when L’Hôpital needs to be applied 3+ times.

Speed shortcut: Memorise the first three terms of standard expansions: sin⁡x≈x−x3/6\sin x \approx x - x^3/6, cos⁡x≈1−x2/2\cos x \approx 1 - x^2/2, ex≈1+x+x2/2e^x \approx 1 + x + x^2/2, ln⁡(1+x)≈x−x2/2\ln(1+x) \approx x - x^2/2. Covers ~80% of JEE limits.

Alternative Method — L’Hôpital’s Rule (3 applications)

Differentiate top and bottom three times:

After 1st: 3cos⁡(3x)−3cos⁡x3x2\frac{3\cos(3x) - 3\cos x}{3x^2} — still 0/00/0.

After 2nd: −9sin⁡(3x)+3sin⁡x6x\frac{-9\sin(3x) + 3\sin x}{6x} — still 0/00/0.

After 3rd: −27cos⁡(3x)+3cos⁡x6\frac{-27\cos(3x) + 3\cos x}{6}.

At x=0x = 0: −27+36=−24/6=−4\frac{-27 + 3}{6} = -24/6 = -4. Same answer.

L’Hôpital works but is slower for higher-order limits.

Common Mistake

Students often apply L’Hôpital once, see another 0/00/0 form, and panic. The rule is iterative — keep differentiating until the limit is determinate.

Another classic: using only the first term of Taylor series (sin⁡x≈x\sin x \approx x) for the numerator. That gives 3x−3x=03x - 3x = 0 — the leading terms cancel, and we MUST go to the next non-zero term (x3x^3).

JEE Main and Advanced both ask this template every year. The trick is recognising “leading terms cancel” — if the obvious approximation gives 00, expand further. Recurring favourite: lim⁡x→0(x−sin⁡x)/x3=1/6\lim_{x \to 0} (x - \sin x)/x^3 = 1/6.

For CBSE Class 11, simpler limits like lim⁡x→0sin⁡x/x=1\lim_{x \to 0} \sin x / x = 1 are tested. Same Taylor logic, fewer terms needed.

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