Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 15 / c

sin x = x - x^{3} /6 + O (x^{5})

and

(1 + x)^{- 2} = 1 - 2 x + 3 x^{2} + O (x^{3})

. Multiplying through the cubic order gives

\frac{sin x}{( 1 + x ) ^{2}} = x - 2 x^{2} + \frac{17}{6} x^{3} + O (x^{4}) .

These are the first three nonzero terms; the constant term vanishes.

ii

Both numerator and denominator vanish quadratically. Factor out

x^{2}

before dividing the power series:

\frac{x s i n x}{l n ( 1 + x ^{2} )} = \frac{1 - x ^{2} /6 + x ^{4} /120 + O ( x ^{6} )}{1 - x ^{2} /2 + x ^{4} /3 + O ( x ^{6} )} .

The reciprocal of the denominator is

1 + x^{2} /2 - x^{4} /12 + O (x^{6})

. Hence

\frac{x sin x}{ln ( 1 + x ^{2} )} = 1 + \frac{x ^{2}}{3} - \frac{19 x ^{4}}{120} + O (x^{6}) .

The expansion describes the continuous extension at the removable singularity

x = 0

, where the original quotient is undefined but its limit is one. Odd powers vanish because the quotient is even.

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