Use the series expansion
$$\ln ( 1 + x ) = x - \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 } + \ldots$$
to write down the first four terms in the expansion, in ascending powers of \(x\), of \(\ln ( 1 - x )\).
The function f is defined by
$$\mathrm { f } ( x ) = \mathrm { e } ^ { \sin x }$$
Use Maclaurin's theorem to show that when \(\mathrm { f } ( x )\) is expanded in ascending powers of \(x\) :
the first three terms are
$$1 + x + \frac { 1 } { 2 } x ^ { 2 }$$
the coefficient of \(x ^ { 3 }\) is zero.
Find
$$\lim _ { x \rightarrow 0 } \frac { \mathrm { e } ^ { \sin x } - 1 + \ln ( 1 - x ) } { x ^ { 2 } \sin x }$$
(4 marks)