- The Taylor series expansion of \(f ( x )\) about \(x = a\) is given by
$$f ( x ) = f ( a ) + ( x - a ) f ^ { \prime } ( a ) + \frac { ( x - a ) ^ { 2 } } { 2 ! } f ^ { \prime \prime } ( a ) + \ldots + \frac { ( x - a ) ^ { r } } { r ! } f ^ { ( r ) } ( a ) + \ldots$$
- (a) Use differentiation to determine the Taylor series expansion of \(\ln x\), in ascending powers of ( \(x - 1\) ), up to and including the term in \(( x - 1 ) ^ { 2 }\)
(b) Hence prove that
$$\lim _ { x \rightarrow 1 } \left( \frac { \ln x } { x - 1 } \right) = 1$$ - Use L'Hospital's rule to determine
$$\lim _ { x \rightarrow 0 } \left( \frac { 1 } { ( x + 3 ) \tan ( 6 x ) \operatorname { cosec } ( 2 x ) } \right)$$
(Solutions relying entirely on calculator technology are not acceptable.)