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

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

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]