91Ó°ÊÓ

f ( x ) = log ( 1 + x ) aroundx = 0

Referring to the Exercise 13 in Sec. 4.2.

\[{\bf{g}}\left( {\bf{x}} \right){\bf{ = g}}\left( {{{\bf{x}}_{\bf{0}}}} \right){\bf{ + }}\left( {{\bf{x - }}{{\bf{x}}_{\bf{0}}}} \right){\bf{g'}}\left( {{{\bf{x}}_{\bf{0}}}} \right){\bf{ + }}\frac{{{{\left( {{\bf{x}} - {{\bf{x}}_{\bf{0}}}} \right)}^{\bf{2}}}}}{{\bf{2}}}{\bf{g''}}\left( {\bf{y}} \right)\]

Where, g', g'' be the derivatives of g ( x ) at x = x₀ and y between x and x₀

Let,\[{\left( {1 + {a_n}} \right)^{{c_n}}}{e^{ - {a_n}{c_n}}} = {e^{\left[ {{c_n}\log \left( {1 + {a_n}} \right) - {a_n}{c_n}} \right]}}\]

Using Taylor’s theorem with remainder,

\[\log \left( {1 + {a_n}} \right) = {a_n} - \frac{{{a_n}^2}}{{2{{\left( {1 + {y_n}} \right)}^2}}}\]; where,\[{y_n}\]between 0 and\[{a_n}\]

It follows,

\[\begin{array}{c}{c_n}\log \left( {1 + {a_n}} \right) - {a_n}{c_n} = {c_n}{a_n} - \frac{{{c_n}{a_n}^2}}{{2{{\left( {1 + {y_n}} \right)}^2}}} - {a_n}{c_n}\\ = - \frac{{{c_n}{a_n}^2}}{{2{{\left( {1 + {y_n}} \right)}^2}}}\end{array}\]

Assume that\[\mathop {\lim }\limits_{n \to \infty } {c_n}{a_n}^2 = 0\].

Since\[{y_n}\]between 0 and\[{a_n}\]\[{a_n}\]goes to 0, then,

\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{{2{{\left( {1 + {y_n}} \right)}^2}}} = 0\]

Therefore,

\[\begin{array}{l}\mathop {\lim }\limits_{n \to \infty } {c_n} - \frac{{{c_n}{a_n}^2}}{{2{{\left( {1 + {y_n}} \right)}^2}}} = 0\\ \Rightarrow \mathop {\lim }\limits_{n \to \infty } {c_n}\log \left( {1 + {a_n}} \right) - {a_n}{c_n} = 0\end{array}\]

Hence, proved.