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  4. If f(x)=( log e(1+x2 tan x)/ sin x3),x≠ 0 is to be continuous at x=0, then f(0) must be defined as

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Q. If $ f(x)=\frac{{{\log }_{e}}(1+{{x}^{2}}\tan x)}{\sin {{x}^{3}}},x\ne 0 $ is to be continuous at $ x=0, $ then $ f(0) $ must be defined as

KEAMKEAM 2007Continuity and Differentiability

Solution:

$ f(x)=\frac{{{\log }_{e}}(1+{{x}^{2}}\tan x)}{\sin {{x}^{3}}} $ This function is continuous at $ x=0, $ then $ \underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}(1+{{x}^{2}}\tan x)}{\sin {{x}^{3}}}=f(0) $
$ \Rightarrow $ $ \underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}\left\{ 1+{{x}^{2}}\left( x+\frac{{{x}^{3}}}{3}+\frac{2{{x}^{5}}}{15}+..... \right) \right\}}{{{x}^{3}}-\frac{{{x}^{9}}}{3!}+\frac{{{x}^{15}}}{5!}-....} $
$ \Rightarrow $ $ \underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}(1+{{x}^{3}})}{{{x}^{3}}-\frac{{{x}^{9}}}{3!}+\frac{{{x}^{15}}}{5!}-....}=f(0) $ [on neglecting higher power of $ x $ in $ {{x}^{2}}tan\text{ }x $ ]
$ \Rightarrow $ $ \underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{3}}-\frac{{{x}^{6}}}{2}+\frac{{{x}^{9}}}{3}-.....}{{{x}^{3}}-\frac{{{x}^{9}}}{3!}+\frac{{{x}^{15}}}{5!}-....}=f(0) $
$ \Rightarrow $ $ 1=f(0) $