Use l’Hôpital’s rule to find \(\mathop {\lim }\limits_{x \to 0} (\csc x - \cot x)\).

\(\mathop {\lim }\limits_{x \to 0} (\csc x - \cot x) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos x}}{{\sin x}}} \right)\)     M1A1

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

\( = 0\)     A1

This question was well answered in general although some of the weaker candidates differentiated the whole expression rather than the numerator and denominator separately. Some candidates wrote \(\cos {\text{ec}}x - \cot x = \frac{1}{{\sin x}} - \frac{1}{{\tan x}} = \frac{{\tan x - \sin x}}{{\sin x\tan x}}\) which is correct but it introduced an extra round of differentiation with opportunity for error.