Find the value of \(\mathop {\lim }\limits_{x \to 1} \left( {\frac{{\ln x}}{{\sin 2\pi x}}} \right)\).

By using the series expansions for \({{\text{e}}^{{x^2}}}\) and cos x evaluate \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - {{\text{e}}^{{x^2}}}}}{{1 - \cos x}}} \right).\).

Using l’Hopital’s rule,

\(\mathop {\lim }\limits_{x \to 1} \left( {\frac{{\ln x}}{{\sin 2\pi x}}} \right) = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{\frac{1}{x}}}{{2\pi \cos 2\pi x}}} \right)\)     M1A1

\( = \frac{1}{{2\pi }}\)     A1

[3 marks]

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - {{\text{e}}^{{x^2}}}}}{{1 - \cos x}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \left( {1 + {x^2} + \frac{{{x^4}}}{{2!}} + \frac{{{x^6}}}{{3!}} + ...} \right)}}{{1 - \left( {1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - ...} \right)}}} \right)\)     M1A1A1

Note: Award M1 for evidence of using the two series.

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { - {x^2} - \frac{{{x^4}}}{{2!}} - \frac{{{x^6}}}{{3!}} - ...} \right)}}{{\left( {\frac{{{x^2}}}{{2!}} - \frac{{{x^4}}}{{4!}} + ...} \right)}}} \right)\)     A1

EITHER

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { - 1 - \frac{{{x^2}}}{{2!}} - \frac{{{x^4}}}{{3!}} - ...} \right)}}{{\left( {\frac{1}{{2!}} - \frac{{{x^2}}}{{4!}} + ...} \right)}}} \right)\)     M1A1

\( = \frac{{ - 1}}{{\frac{1}{2}}} = - 2\)     A1

OR

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { - 2x - \frac{{4{x^3}}}{{2!}} - \frac{{6{x^5}}}{{3!}} - ...} \right)}}{{\left( {\frac{{2x}}{{2!}} - \frac{{4{x^3}}}{{4!}} + ...} \right)}}} \right)\)     M1A1

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( { - 2 - \frac{{4{x^2}}}{{2!}} - \frac{{6{x^4}}}{{3!}} - ...} \right)}}{{\left( {1 - \frac{{4{x^2}}}{{4!}} + ...} \right)}}} \right)\)

\( = \frac{{ - 2}}{1} = - 2\)     A1

[7 marks]

Part (a) was well done but too often the instruction to use series in part (b) was ignored. When this hint was observed correct solutions followed. 

Part (a) was well done but too often the instruction to use series in part (b) was ignored. When this hint was observed correct solutions followed.