Which one of the following is correct in respect of the function $$f\left( x \right) = \left| x \right| + {x^2}$$

Solution :
$$\because \,f\left( x \right) = \left| x \right| + {x^2}$$
\[ \Rightarrow \,f\left( x \right) = \left\{ \begin{array}{l} {x^2} + x,\,\,\,\,\,x \ge 0\\ {x^2} - x,\,\,\,\,\,x < 0 \end{array} \right.\]
$$\eqalign{ & {\text{L}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) \cr & = \mathop {\lim }\limits_{h \to 0} f\left( {0 - h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} {\left( {0 - h} \right)^2} - \left( {0 - h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} {h^2} + h \cr & = 0 \cr & {\text{and}} \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{x \to 0} f\left( x \right) \cr & = \mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} {\left( {0 + h} \right)^2} + \left( {0 + h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} {h^2} + h \cr & = 0 \cr & \Rightarrow {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = f\left( 0 \right) \cr & \Rightarrow \,f\left( x \right){\text{ is continuous at }}x = 0 \cr & {\text{Now, L}}{\text{.H}}{\text{.D}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} + h}}{{ - h}} \cr & = - \mathop {\lim }\limits_{h \to 0} \,h + 1 \cr & = - 1 \cr & {\text{and, R}}{\text{.H}}{\text{.D}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} + h}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \,h + 1 \cr & = 1 \cr & {\text{Thus, L}}{\text{.H}}{\text{.D}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} \cr & \Rightarrow \,f\left( x \right){\text{ is not differentiable at }}x = 0 \cr} $$