Using the result \(\mathop {{\text{lim}}}\limits_{t \to 0} \frac{{\sin t}}{t} = 1\), or otherwise, calculate \(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{g(2x) - g(3x)}}{{4{x^2}}}\).

Use the Maclaurin series of \(\sin x\) to show that \(g(x) = \sum\limits_{n = 0}^\infty  {{{( - 1)}^n}\frac{{{x^{4n + 2}}}}{{(2n + 1)!}}} \)

Hence determine the minimum number of terms of the expansion of \(g(x)\) required to approximate the value of \(\int_0^1 {g(x){\text{d}}x} \) to four decimal places.

METHOD 1

\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\sin 4{x^2} - \sin 9{x^2}}}{{4{x^2}}}\)     M1

\( = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\sin 4{x^2}}}{{4{x^2}}} - \frac{9}{4}\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\sin 9{x^2}}}{{9{x^2}}}\)     A1A1

\( = 1 - \frac{9}{4} \times 1 =  - \frac{5}{4}\)     A1

METHOD 2

\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\sin 4{x^2} - \sin 9{x^2}}}{{4{x^2}}}\)     M1

\( = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{8x\cos 4{x^2} - 18x\cos 9{x^2}}}{{8x}}\)     M1A1

\( = \frac{{8 - 18}}{8} =  - \frac{{10}}{8} =  - \frac{5}{4}\)     A1

[4 marks]

since \(\sin x = \sum\limits_{n = 0}^\infty  {{{( - 1)}^n}\frac{{{x^{(2n + 1)}}}}{{(2n + 1)!}}} \)   \(\left( {{\text{or }}\sin x = \frac{x}{{1!}} - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} -  \ldots } \right)\)     (M1)

\(\sin {x^2} = \sum\limits_{n = 0}^\infty  {{{( - 1)}^n}\frac{{{x^{2(2n + 1)}}}}{{(2n + 1)!}}} \)   \(\left( {{\text{or }}\sin x = \frac{{{x^2}}}{{1!}} - \frac{{{x^6}}}{{3!}} + \frac{{{x^{10}}}}{{5!}} -  \ldots } \right)\)     A1

\(g(x) = \sin {x^2} = \sum\limits_{n = 0}^\infty  {{{( - 1)}^n}\frac{{{x^{4n + 2}}}}{{(2n + 1)!}}} \)     AG

[2 marks]

let \(I = \int_0^1 {\sin {x^2}{\text{d}}x} \)

\( = \sum\limits_{n = 0}^\infty  {{{( - 1)}^n}\frac{1}{{(2n + 1)!}}} \int_0^1 {{x^{4n + 2}}{\text{d}}x{\text{ }}\left( {\int_0^1 {\frac{{{x^2}}}{{1!}}{\text{d}}x - } \int_0^1 {\frac{{{x^6}}}{{3!}}{\text{d}}x + } \int_0^1 {\frac{{{x^{10}}}}{{5!}}{\text{d}}x -  \ldots } } \right)} \)     M1

\( = \sum\limits_{n = 0}^\infty  {{{( - 1)}^n}\frac{1}{{(2n + 1)!}}} \frac{{[{x^{4n + 3}}]_0^1}}{{(4n + 3)}}{\text{ }}\left( {\left[ {\frac{{{x^3}}}{{3 \times 1!}}} \right]_0^1 - \left[ {\frac{{{x^7}}}{{7 \times 3!}}} \right]_0^1 + \left[ {\frac{{{x^{11}}}}{{11 \times 5!}}} \right]_0^1 -  \ldots } \right)\)     M1

\( = \sum\limits_{n = 0}^\infty  {{{( - 1)}^n}\frac{1}{{(2n + 1)!(4n + 3)}}} {\text{ }}\left( {\frac{1}{{3 \times 1!}} - \frac{1}{{7 \times 3!}} + \frac{1}{{11 \times 5!}} -  \ldots } \right)\)     A1

\( = \sum\limits_{n = 0}^\infty  {{{( - 1)}^n}{a_n}} \) where \({a_n} = \frac{1}{{(4n + 3)(2n + 1)!}} > 0\) for all \(n \in \mathbb{N}\)

as \(\{ {a_n}\} \) is decreasing the sum of the alternating series \(\sum\limits_{n = 0}^\infty  {{{( - 1)}^n}{a_n}} \)

lies between \(\sum\limits_{n = 0}^N {{{( - 1)}^n}{a_n}} \) and \(\sum\limits_{n = 0}^N {{{( - 1)}^n}{a_n}}  \pm {a_{N + 1}}\)     R1

hence for four decimal place accuracy, we need \(\left| {{a_{N + 1}}} \right| < 0.00005\)     M1

since \({a_{2 + 1}} < 0.00005\)     R1

so \(N = 2\)   (or 3 terms)     A1

[7 marks]

Part (a) of this question was accessible to the vast majority of candidates, who recognised that L’Hôpital’s rule could be used. Most candidates were successful in finding the limit, with some making calculation errors. Candidates that attempted to use \(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\sin x}}{x} = 1\) or a combination of this result and L’Hôpital’s rule were less successful.

In part (b) most candidates showed to be familiar with the substitution given and were successful in showing the result.

Very few candidates were able to do part (c) successfully. Most used trial and error to arrive at the answer.