Limits and Derivatives: Diagram-Based Questions (1)

easy • 2 min read
Tags Limits And Derivatives

Question

Evaluate lim⁡x→0sin⁡3x−sin⁡2xx\displaystyle\lim_{x \to 0} \frac{\sin 3x - \sin 2x}{x}.

Solution — Step by Step

lim⁡x→0sin⁡3x−sin⁡2xx=lim⁡x→0sin⁡3xx−lim⁡x→0sin⁡2xx\displaystyle\lim_{x \to 0} \frac{\sin 3x - \sin 2x}{x} = \lim_{x \to 0} \frac{\sin 3x}{x} - \lim_{x \to 0} \frac{\sin 2x}{x}.

(Limits of differences split when each limit exists.)

Standard result: lim⁡x→0sin⁡kxx=k\displaystyle\lim_{x \to 0} \frac{\sin kx}{x} = k.

So lim⁡x→0sin⁡3xx=3\displaystyle\lim_{x \to 0} \frac{\sin 3x}{x} = 3 and lim⁡x→0sin⁡2xx=2\displaystyle\lim_{x \to 0} \frac{\sin 2x}{x} = 2.

lim⁡x→0sin⁡3x−sin⁡2xx=3−2=1\displaystyle\lim_{x \to 0} \frac{\sin 3x - \sin 2x}{x} = 3 - 2 = 1.

Final answer: 1\mathbf{1}.

Why This Works

The standard limit lim⁡x→0sin⁡x/x=1\lim_{x \to 0} \sin x/x = 1 generalizes to lim⁡x→0sin⁡(kx)/x=k\lim_{x \to 0} \sin(kx)/x = k via the substitution u=kxu = kx: sin⁡(kx)/x=k⋅sin⁡(u)/u→k⋅1=k\sin(kx)/x = k \cdot \sin(u)/u \to k \cdot 1 = k.

The split is legal because both individual limits exist. If they didn’t, we’d need a different approach (like L’Hopital or Taylor expansion).

Alternative Method

Apply L’Hopital (since this is 0/00/0): lim⁡x→03cos⁡3x−2cos⁡2x1=3cos⁡0−2cos⁡0=3−2=1\lim_{x \to 0} \frac{3\cos 3x - 2\cos 2x}{1} = 3\cos 0 - 2\cos 0 = 3 - 2 = 1. Same answer in one line for a calculus-savvy student.

Or use the sum-to-product identity: sin⁡3x−sin⁡2x=2cos⁡(5x/2)sin⁡(x/2)\sin 3x - \sin 2x = 2\cos(5x/2)\sin(x/2), then lim⁡2cos⁡(5x/2)⋅sin⁡(x/2)/x=2⋅1⋅12=1\lim 2\cos(5x/2) \cdot \sin(x/2)/x = 2 \cdot 1 \cdot \frac{1}{2} = 1.

Common Mistake

A frequent slip: students cancel xx between numerator and denominator naively, getting sin⁡3−sin⁡2\sin 3 - \sin 2 — completely wrong. You can’t cancel xx from inside sin⁡(kx)\sin(kx). Always reduce to known forms (sin⁡kx/x\sin kx/x or use L’Hopital) before evaluating.

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