Let $$f\left( x \right) = \left[ {{{\tan }^2}x} \right],$$    where $$\left[ . \right]$$ denotes the greatest integer function. Then :

Solution :
$$\eqalign{ & \mathop {\lim }\limits_{x \to 0 + 0} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \left[ {{{\tan }^2}h} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \,0 = 0 \cr & \mathop {\lim }\limits_{x \to 0 - 0} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {0 - h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \left[ {{{\tan }^2}\left( { - h} \right)} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \,0 = 0\,;\,f\left( 0 \right) = 0 \cr & \therefore \mathop {\lim }\limits_{h \to 0} f\left( x \right) = f\left( 0 \right) = 0 \cr & \therefore f\left( x \right){\text{ continuous at }}x = 0 \cr & \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\tan }^2}h} \right] - 0}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{0}{h} = \mathop {\lim }\limits_{h \to 0} \,0 = 0 \cr & \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\tan }^2}\left( { - h} \right)} \right] - 0}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{0}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \,0 = 0 \cr & {\text{So, }}f'\left( 0 \right) = 0 \cr} $$