Find the value of \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \cot x} \right)\) .

Find the interval of convergence of the infinite series\[\frac{{(x + 2)}}{{3 \times 1}} + \frac{{{{(x + 2)}^2}}}{{{3^2} \times 2}} + \frac{{{{(x + 2)}^3}}}{{{3^3} \times 3}} +  \ldots \]

(i)     Find the Maclaurin series for \(\ln (1 + \sin x)\) up to and including the term in \({x^3}\) .

(ii)     Hence find a series for \(\ln (1 - \sin x)\) up to and including the term in \({x^3}\) .

(iii)     Deduce, by considering the difference of the two series, that \(\ln 3 \simeq \frac{\pi }{3}\left( {1 + \frac{{{\pi ^2}}}{{216}}} \right)\) .

EITHER

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \cot x} \right)\)

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan x - x}}{{x\tan x}}} \right)\)     M1A1

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{{\sec }^2}x - 1}}{{x{{\sec }^2}x + \tan x}}} \right)\) , using l’Hopital     A1

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{2{{\sec }^2}x\tan x}}{{2{{\sec }^2}x + 2x{{\sec }^2}x\tan x}}} \right)\)     A1A1

\( = 0\)     A1

OR

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \cot x} \right)\)

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x - x\cos x}}{{x\sin x}}} \right)\)     M1A1

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x\sin x}}{{\sin x + x\cos x}}} \right)\) , using l’Hopital     A1

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x + x\cos x}}{{2\cos x - x\sin x}}} \right)\)     A1A1

\( = 0\)     A1

[6 marks]

\({u_n} = \frac{{{{(x + 2)}^n}}}{{{3^n} \times n}}\)     A1

\(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{\frac{{{{(x + 2)}^{n + 1}}}}{{{3^{n + 1}} \times (n + 1)}}}}{{\frac{{{{(x + n)}^n}}}{{{3^n} \times n}}}} = \frac{{(x + 2)n}}{{3(n + 1)}}\)     M1A1

\(\mathop {\lim }\limits_{n \to \infty } \frac{{(x + 2)n}}{{3(n + 1)}} = \frac{{(x + 2)}}{3}\)     M1A1

\(\left| {\frac{{(x + 2)}}{3}} \right| < 1 \Rightarrow  - 5 < x < 1\)     M1A1

if \(x = 1\) series is \(1 + \frac{1}{2} + \frac{1}{3} +  \ldots \) which diverges     A1

if \(x = - 5\) series is \( - 1 + \frac{1}{2} - \frac{1}{3} +  \ldots  + \frac{{{{( - 1)}^n}}}{n}\) which converges     A1

hence interval is \( - 5 \le x < 1\)     A1

[10 marks]

(i)     \(f(x) = \ln (1 + \sin x)\) , \(f(0) = 0\)     A1

\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\) , \(f'(0) = 1\)     A1

\(f''(x) = \frac{{ - \sin x(1 + \sin x) - {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}} = \frac{{ - (1 + \sin x)}}{{{{(1 + \sin x)}^2}}} = \frac{{ - 1}}{{1 + \sin x}}\) , \(f''(0) =  - 1\)     A1

\(f'''(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\) , \(f'''(0) = 1\)     A1

\(\ln (1 + \sin x) \approx x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} -  \ldots \)     A1

(ii)     \( - \sin x = \sin ( - x)\)    M1

so, \(\ln (1 - \sin x) \approx  - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{6} -  \ldots \)     A1

(iii)     \(\ln (1 + \sin x) - ln(1 - \sin x)\)

\( = \ln \left( {\frac{{1 + \sin x}}{{1 - \sin x}}} \right) \approx 2x + \frac{{{x^3}}}{3}\)     M1A1

let \(x = \frac{\pi }{6}\) then, \(\ln \left( {\frac{{1 + \frac{1}{2}}}{{1 - \frac{1}{2}}}} \right) = \ln 3 \approx 2\left( {\frac{\pi }{6}} \right) + \frac{{{{\left( {\frac{\pi }{6}} \right)}^3}}}{3}\)     M1A1A1

\( = \frac{\pi }{3}\left( {1 + \frac{{{\pi ^2}}}{{216}}} \right)\)     AG

[12 marks]

There was some confusion in differentiating twice using l’Hopital’s Rule but the confusion was made worse by not taking care to write legibly.

This was in general well done but some students did not bother to test the end points.

(c)(i) This was generally well done with various approaches being used.

(ii) This part was often done by using the differentiation all over again instead of using part (i) again demonstrating a lack of appreciation of where time and effort can be saved in answering questions and ignoring the word "Hence".

(iii) Candidates often managed to work their way through this question but with lack of clarity as to where \(\frac{p}{6}\) came from.