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]