Find the values of ${a_0},{\text{ }}{a_1},{\text{ }}{a_2}$ and ${a_3}$.

Hence, or otherwise, find the value of $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{x - 1}}$.

let $f(x) = \sqrt x ,{\text{ }}f(1) = 1$     (A1)

$f'(x) = \frac{1}{2}{x^{ - \frac{1}{2}}},{\text{ }}f'(1) = \frac{1}{2}$     (A1)

$f''(x) = - \frac{1}{4}{x^{ - \frac{3}{2}}},{\text{ }}f''(1) = - \frac{1}{4}$     (A1)

$f'''(x) = \frac{3}{8}{x^{ - \frac{5}{2}}},{\text{ }}f'''(1) = \frac{3}{8}$     (A1)

${a_1} = \frac{1}{2} \cdot \frac{1}{{1!}},{\text{ }}{a_2} = - \frac{1}{4} \cdot \frac{1}{{2!}},{\text{ }}{a_3} = \frac{3}{8} \cdot \frac{1}{{3!}}$     (M1)

${a_0} = 1,{\text{ }}{a_1} = \frac{1}{2},{\text{ }}{a_2} = - \frac{1}{8},{\text{ }}{a_3} = \frac{1}{{16}}$     A1

Note: Accept $y = 1 + \frac{1}{2}(x - 1) - \frac{1}{8}{(x - 1)^2} + \frac{1}{{16}}{(x - 1)^3} +  \ldots$

[6 marks]

METHOD 1

$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2}(x - 1) - \frac{1}{8}{{(x - 1)}^2} +  \ldots }}{{x - 1}}$     M1

$= \mathop {\lim }\limits_{x \to 1} \left( {\frac{1}{2} - \frac{1}{8}(x - 1) +  \ldots } \right)$     A1

$= \frac{1}{2}$     A1

METHOD 2

using l’Hôpital’s rule,     M1

$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2}{x^{ - \frac{1}{2}}}}}{1}$     A1

$= \frac{1}{2}$     A1

METHOD 3

$\frac{{\sqrt x  - 1}}{{x + 1}} = \frac{1}{{\sqrt x  + 1}}$     M1A1

$\mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x  + 1}} = \frac{1}{2}$     A1

[3 marks]

Many candidates achieved full marks on this question but there were still a large minority of candidates who did not seem familiar with the application of Taylor series. Whilst all candidates who responded to this question were aware of the need to use derivatives many did not correctly use factorials to find the required coefficients. It should be noted that the formula for Taylor series appears in the Information Booklet.

Many candidates achieved full marks on this question but there were still a large minority of candidates who did not seem familiar with the application of Taylor series. Whilst all candidates who responded to this question were aware of the need to use derivatives many did not correctly use factorials to find the required coefficients. It should be noted that the formula for Taylor series appears in the Information Booklet.