Use L’Hôpital’s Rule to find \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} - 1 - x\cos x}}{{{{\sin }^2}x}}\) .

apply l’Hôpital’s Rule to a \(0/0\) type limit

\(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} - 1 - x\cos x}}{{{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} - \cos x + x\sin x}}{{2\sin x\cos x}}\)     M1A1

noting this is also a \(0/0\) type limit, apply l’Hôpital’s Rule again     (M1)

obtain \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^x} + \sin x + x\cos x + \sin x}}{{2\cos 2x}}\)     A1

substitution of x = 0     (M1)

= 0.5     A1

[6 marks]

The vast majority of candidates were familiar with L’Hôpitals rule and were also able to apply the technique twice as required by the problem. The errors that occurred were mostly due to difficulty in applying the differentiation rules correctly or errors in algebra. A small minority of candidates tried to use the quotient rule but it seemed that most candidates had a good understanding of L’Hôpital’s rule and its application to finding a limit.