20N.1.AHL.TZ0.F_1

pestleMathematicsAAHLPaper 120N· ahl-5-13-limits-and-lhopitalssource ↗

Use l’Hôpital’s rule to determine the value of

$\lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}}$ .

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

using l’Hôpital’s rule,

$\lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}} = \lim_{x \to 0} \frac{2 \cos x - 2 \cos 2 x}{3 x^{2}}$ M1A1

$= \lim_{x \to 0} \frac{- 2 \sin x + 4 \sin 2 x}{6 x}$ (M1)A1

$= \lim_{x \to 0} \frac{- 2 \cos x + 8 \cos 2 x}{6}$ A1

$= 1$ A1

[6 marks]

Examiners’ report

[N/A]