Find $\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{x + {x^2}}}$ ;

Find $\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^2} + 2{x^2}\ln x}}{{1 - \sin \frac{{\pi x}}{2}}}$ .

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{x + {x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sec }^2}x}}{{1 + 2x}}$     M1A1A1

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{x + {x^2}}} = \frac{1}{1} = 1$     A1

[4 marks] 

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^2} + 2{x^2}\ln x}}{{1 - \sin \frac{{\pi x}}{2}}} = \mathop {\lim }\limits_{x \to 1} \frac{{ - 2x + 2x + 4x\ln x}}{{ - \frac{\pi }{2}\cos \frac{{\pi x}}{2}}}$     M1A1A1

$= \mathop {\lim }\limits_{x \to 1} \frac{{4 + 4\ln x}}{{\frac{{{\pi ^2}}}{4}\sin \frac{{\pi x}}{2}}}$     M1A1A1

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^2} + 2{x^2}\ln x}}{{1 - \sin \frac{{\pi x}}{2}}} = \frac{4}{{\frac{{{\pi ^2}}}{4}}} = \frac{{16}}{{{\pi ^2}}}$     A1

[7 marks]

This question was accessible to the vast majority of candidates, who recognised that L’Hopital’s rule was required. A few of the weaker candidates did not realise that it needed to be applied twice in part (b). Many fully correct solutions were seen.

This question was accessible to the vast majority of candidates, who recognised that L’Hopital’s rule was required. A few of the weaker candidates did not realise that it needed to be applied twice in part (b). Many fully correct solutions were seen.