Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 20 / b

sin (3 x) = 3 x + O (x^{3})

and

sinh x = x + O (x^{3})

. Dividing by

x

in a punctured neighborhood of zero gives

x \to 0 lim \frac{sin ( 3 x )}{sinh x} = 3.

ii

The linear terms cancel:

tan x = x + x^{3} /3 + O (x^{5})

, whereas

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

. Also

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

. Their leading nonzero terms therefore give

x \to 0 lim \frac{tan x - tanh x}{sinh x - x} = \frac{2/3}{1/6} = 4.

The Maclaurin series remainder orders justify division after factoring out

x^{3}

To retain the finite term after cancellation, expand far enough before dividing. The Maclaurin series are

csc x = x^{- 1} + \frac{x}{6} + \frac{7 x ^{3}}{360} + O (x^{5}), sinh x = x + \frac{x ^{3}}{6} + \frac{x ^{5}}{120} + O (x^{7}) .

Thus

\frac{c s c x}{x ^{3}} - \frac{s i n h x}{x ^{5}} = (x^{- 4} + \frac{x ^{- 2}}{6} + \frac{7}{360}) - (x^{- 4} + \frac{x ^{- 2}}{6} + \frac{1}{120}) + O (x^{2}),

and hence

x \to 0 lim (\frac{csc x}{x ^{3}} - \frac{sinh x}{x ^{5}}) = \frac{1}{90} .

There are no discussions about this article yet.

There are currently no matching articles.