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21M.1.AHL.TZ1.8

pestleMathematicsAAHLPaper 121M· ahl-5-13-limits-and-lhopitalssource ↗

Use l’Hôpital’s rule to find limx0arctan2xtan3x.

Markscheme / solution

attempt to differentiate numerator and denominator        M1

limx0arctan2xtan3x

=limx021+4x23sec23x        A1A1

 

Note: A1 for numerator and A1 for denominator. Do not condone absence of limits.

 

attempt to substitute x=0         (M1)

=23        A1

 

Note: Award a maximum of M1A1A0M1A1 for absence of limits.

 

[5 marks]

Examiners’ report
[N/A]