The function f is defined by $f(x) = {{\text{e}}^{({{\text{e}}^x} - 1)}}$ .

(a)     Assuming the Maclaurin series for ${{\text{e}}^x}$ , show that the Maclaurin series for $f(x)$

is $1 + x + {x^2} + \frac{5}{6}{x^3} + \ldots {\text{ .}}$

(b)     Hence or otherwise find the value of $\mathop {\lim }\limits_{x \to 0} \frac{{f(x) - 1}}{{f'(x) - 1}}$ .

(a)     ${{\text{e}}^x} - 1 = x + \frac{{{x^2}}}{2} + \frac{{{x^2}}}{6} + \ldots$     A1

${{\text{e}}^{{{\text{e}}^x} - 1}} = 1 + \left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right) + \frac{{{{\left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right)}^2}}}{2} + \frac{{{{\left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right)}^3}}}{6} + \ldots$     M1A1

$= 1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{2} + \frac{{{x^3}}}{6} + \ldots$     M1A1

$= 1 + x + {x^2} + \frac{5}{6}{x^3} + \ldots$     AG

[5 marks]

(b)     EITHER

$f'(x) = 1 + 2x + \frac{{5{x^2}}}{2} + \ldots$     A1

$\frac{{f(x) - 1}}{{f'(x) - 1}} = \frac{{x + {x^2} + 5{x^3}/6 + \ldots }}{{2x + 5{x^2}/2 + \ldots }}$     M1A1

$= \frac{{1 + x + \ldots }}{{2 + 5x/2 + \ldots }}$     A1

$\to \frac{1}{2}{\text{ as }}x \to 0$     A1

[5 marks]

OR

using l’Hopital’s rule,     M1

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

$= \mathop {\lim }\limits_{x \to 0} \frac{{{{\text{e}}^{({{\text{e}}^x} + x - 1)}}}}{{{{\text{e}}^{({{\text{e}}^x} + x - 1)}} \times ({{\text{e}}^x} + 1)}}$     A1

$= \frac{1}{2}$     A1

[5 marks]

Total [10 marks]

Many candidates obtained the required series by finding the values of successive derivatives at x = 0 , failing to realise that the intention was to start with the exponential series and replace x by the series for ${{\text{e}}^x} - 1$. Candidates who did this were given partial credit for using this method. Part (b) was reasonably well answered using a variety of methods.