Find \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos {x^6}}}{{{x^{12}}}}} \right)\).

METHOD 1

\(f(0) = \frac{0}{0}\), hence using l’Hôpital’s Rule,     (M1)

\(g(x) = 1 - \cos ({x^6}),{\text{ }}h(x) = {x^{12}};{\text{ }}\frac{{g'(x)}}{{h'(x)}} = \frac{{6{x^5}\sin ({x^6})}}{{12{x^{11}}}} = \frac{{\sin ({x^6})}}{{2{x^6}}}\)     A1A1

EITHER

\(\frac{{g'(0)}}{{h'(0)}} = \frac{0}{0}\), using l’Hôpital’s Rule again,     (M1)

\(\frac{{g''(x)}}{{h''(x)}} = \frac{{6{x^5}\cos ({x^6})}}{{12{x^5}}} = \frac{{\cos ({x^6})}}{2}\)     A1A1

\(\frac{{g''(0)}}{{h''(0)}} = \frac{1}{2}\), hence the limit is \(\frac{1}{2}\)     A1

OR

So \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos {x^6}}}{{{x^{12}}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}}\)     A1

\( = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}}\)     A1

\( = \frac{1}{2}{\text{ since }}\mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^6}}}{{2{x^6}}} = 1\)     A1 (R1)

METHOD 2

substituting \({{x^6}}\) for x in the expansion \(\cos x = 1 - \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} \ldots \)     (M1)

\(\frac{{1 - \cos {x^6}}}{{{x^{12}}}} = \frac{{1 - \left( {1 - \frac{{{x^{12}}}}{2} + \frac{{{x^{24}}}}{{24}}} \right) \ldots }}{{{x^{12}}}}\)     M1A1

\( = \frac{1}{2} - \frac{{{x^{12}}}}{{24}} + ...\)     A1A1

\(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos {x^6}}}{{{x^{12}}}} = \frac{1}{2}\)     M1A1

Note: Accept solutions using Maclaurin expansions.

[7 marks]

Surprisingly, some weaker candidates were more successful in answering this question than stronger candidates. If candidates failed to simplify the expression after the first application of L’Hôpital’s rule, they generally were not successful in correctly differentiating the expression a \({2^{{\text{nd}}}}\) time, hence could not achieve the final three A marks.