17M.3.AHL.TZ0.HCA_1

pestleMathematicsAAHLPaper 317M· ahl-5-13-limits-and-lhopitalssource ↗

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

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{x\ln (1 + x)}}.$

Markscheme / solution

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

attempt to use l’Hôpital’s rule, M1

${\text{limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{2\sin x\cos x}}{{\ln (1 + x) + \frac{x}{{1 + x}}}}$ $\,\,\,$ or $\,\,\,$ $\frac{{\sin 2x}}{{\ln (1 + x) + \frac{x}{{1 + x}}}}$ A1A1

Note: Award A1 for numerator A1 for denominator.

this gives 0/0 so use the rule again (M1)

$= \mathop {\lim }\limits_{x \to 0} \frac{{2{{\cos }^2}x - 2{{\sin }^2}x}}{{\frac{1}{{1 + x}} + \frac{{1 + x - x}}{{{{(1 + x)}^2}}}}}$ $\,\,\,$ or $\,\,\,$ $\frac{{2\cos 2x}}{{\frac{{2 + x}}{{{{(1 + x)}^2}}}}}$ A1A1

Note: Award A1 for numerator A1 for denominator.

$= 1$ A1

Note: This A1 is dependent on all previous marks being awarded, except when the first application of L’Hopital’s does not lead to 0/0, when it should be awarded for the correct limit of their derived function.

[7 marks]

Examiners’ report

[N/A]