Date | May 2017 | Marks available | 7 | Reference code | 17M.3.AHL.TZ0.Hca_1 |
Level | Additional Higher Level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Determine | Question number | Hca_1 | Adapted from | N/A |
Question
Use l’Hôpital’s rule to determine the value of
limx→0sin2xxln(1+x).
Markscheme
* 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
limit=limx→02sinxcosxln(1+x)+x1+xorsin2xln(1+x)+x1+x A1A1
Note: Award A1 for numerator A1 for denominator.
this gives 0/0 so use the rule again (M1)
=limx→02cos2x−2sin2x11+x+1+x−x(1+x)2or2cos2x2+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]