The function f is defined by
$$\mathrm { f } ( x ) = \ln \left( 1 + \mathrm { e } ^ { x } \right)$$
Use Maclaurin's theorem to show that when \(\mathrm { f } ( x )\) is expanded in ascending powers of \(x\) :
the first three terms are
$$\ln 2 + \frac { 1 } { 2 } x + \frac { 1 } { 8 } x ^ { 2 }$$
the coefficient of \(x ^ { 3 }\) is zero.
Hence write down the first two non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln \left( \frac { 1 + \mathrm { e } ^ { x } } { 2 } \right)\).
Use the series expansion
$$\ln ( 1 + x ) = x - \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } - \ldots$$
to write down the first three terms in the expansion, in ascending powers of \(x\), of \(\ln \left( 1 - \frac { x } { 2 } \right)\).
Use your answers to parts (b) and (c) to find
$$\lim _ { x \rightarrow 0 } \left[ \frac { \ln \left( \frac { 1 + \mathrm { e } ^ { x } } { 2 } \right) + \ln \left( 1 - \frac { x } { 2 } \right) } { x - \sin x } \right]$$