If \[f\left( x \right) = \left\{ \begin{array}{l} \left( {\frac{{{x^2}}}{a}} \right) - a,\,\,\,\,\,{\rm{when\,\, }}x < a\\ \,\,\,\,\,\,\,\,\,\,\,0,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when\,\, }}x = a\\ a - \left( {\frac{{{x^2}}}{a}} \right),\,\,\,\,{\rm{when\,\, }}x > a \end{array} \right.\]        ,then :

Solution :
$$\eqalign{ & f\left( a \right) = 0 \cr & \mathop {\lim }\limits_{x \to a - } f\left( x \right) = \mathop {\lim }\limits_{x \to a - } \left( {\frac{{{x^2}}}{a} - a} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{{{\left( {a - h} \right)}^2}}}{a} - a} \right\} = 0 \cr & {\text{and }}\mathop {\lim }\limits_{x \to a + } f\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {a - \frac{{{{\left( {a + h} \right)}^2}}}{a}} \right\} = 0 \cr} $$
Hence it is continuous at $$x = a.$$