By evaluating successive derivatives at \(x = 0\) , find the Maclaurin series for \(\ln \cos x\) up to and including the term in \({x^4}\) .

Consider \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \cos x}}{{{x^n}}}\) , where \(n \in \mathbb{R}\) .

Using your result from (a), determine the set of values of \(n\) for which

  (i)     the limit does not exist;

  (ii)     the limit is zero;

  (iii)     the limit is finite and non-zero, giving its value in this case.

attempt at repeated differentiation (at least 2)     M1

let \(f(x) = \ln \cos x\) , \(f(0) = 0\)     A1

\(f'(x) = - \tan x\) , \(f'(0) = 0\)     A1

\(f''(x) = - {\sec ^2}x\) , \(f''(0) =  - 1\)     A1

\(f'''(x) = - 2{\sec ^2}x\tan x\) , \(f'''(0) = 0\)     A1

\({f^{iv}}(x) = - 2se{c^4}x - 4se{c^2}xta{n^2}x\) , \({f^{iv}}(0) =  - 2\)       A1

the Maclaurin series is

\(\ln \cos x = - \frac{{{x^2}}}{2} - \frac{{{x^4}}}{{12}} +  \ldots \)     M1A1

Note: Allow follow-through on final A1.

[8 marks]

\(\frac{{\ln \cos x}}{{{x^n}}} = - \frac{{{x^{2 - n}}}}{2} - \frac{{{x^{4 - n}}}}{{12}} +  \ldots \)     (M1)

(i)     the limit does not exist if \(n > 2\)     A1

(ii)     the limit is zero if \(n < 2\)     A1

(iii)     if \(n = 2\) , the limit is \( - \frac{1}{2}\)     A1A1

[5 marks]

A reasonable number of candidates achieved full marks on this question. However, in part (a) a number of candidates struggled to find the Maclaurin series accurately. It was not uncommon to see errors in finding the higher derivatives, which was often caused by not simplifying the answer for earlier derivatives. At this level, it is expected that candidates understand the importance of using the most efficient methods.

A pleasing number of candidates made significant progress or achieved full marks in part (b), provided that they realised the importance of recognizing that\[\frac{{\ln \cos x}}{{{x^n}}} =  - \frac{{{x^{2 - n}}}}{2} - \frac{{{x^{4 - n}}}}{{12}} +  \ldots \]