If \[f\left( x \right) = \left\{ \begin{array}{l} \frac{{x\,\log \,\cos \,x}}{{\log \left( {1 + {x^2}} \right)}},\,\,\,\,\,x \ne 0\\ \,\,\,\,\,\,\,\,\,\,\,0,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \end{array} \right.\]       then $$f\left( x \right)$$  is :

Solution :
We have,
$$\eqalign{ & {\text{L}}f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - h\,\log \,\cos \,h}}{{ - h\,\log \left( {1 + {h^2}} \right)}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\log \,\cos \,h}}{{\log \left( {1 + {h^2}} \right)}}\,\,\,\,\,\,\,\,\left( {\frac{0}{0}{\text{form}}} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - \tan \,h}}{{\frac{{2h}}{{\left( {1 + {h^2}} \right)}}}} \cr & = - \frac{1}{2} \cr & {\text{R}}f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{h\,\log \,\cos \,h}}{{h\,\log \left( {1 + {h^2}} \right)}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\log \,\cos \,h}}{{\log \left( {1 + {h^2}} \right)}}\,\,\,\,\,\,\,\,\left( {\frac{0}{0}{\text{form}}} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - \tan \,h}}{{\frac{{2h}}{{\left( {1 + {h^2}} \right)}}}} \cr & = \frac{{ - 1}}{2} \cr} $$
Since $${\text{L}}f'\left( 0 \right){\text{ = R}}f'\left( 0 \right),$$    therefore $$f\left( x \right)$$  is differentiable at $$x = 0.$$
Since differentiability $$ \Rightarrow $$ continutity, therefore $$f\left( x \right)$$  iscontinuous at $$x = 0.$$