\[\begin{array}{l} {\rm{Let\, }}f\left( x \right) = \left\{ \begin{array}{l} \sin \,x,\,x \ne n\pi \\ 2,\,x = n\pi ,\,n \in Z, \end{array} \right.\\ {\rm{and\, }}g\left( x \right) = \left\{ \begin{array}{l} {x^2} + 1,\,x \ne 2\\ 3,\,x = 2\,\, \end{array} \right. \end{array}\]
Then $$\mathop {\lim }\limits_{x \to 0} g\left( {f\left( x \right)} \right)$$   is :

Solution :
\[\begin{array}{l} g\left\{ {f\left( x \right)} \right\} = \left\{ \begin{array}{l} {\left\{ {f\left( x \right)} \right\}^2} + 1,\,f\left( x \right) \ne 2\\ 3,\,f\left( x \right) = 2 \end{array} \right.\\ \therefore g\left\{ {f\left( x \right)} \right\} = \left\{ \begin{array}{l} {\sin ^2}x + 1,\,x \ne n\pi \\ 3,\,x = n\pi \end{array} \right. \end{array}\]
$$\eqalign{ & {\text{RH limit}} = \mathop {\lim }\limits_{x \to 0 + 0} g\left\{ {f\left( x \right)} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} g\left\{ {f\left( {0 + h} \right)} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} \left\{ {{{\sin }^2}\left( {0 + h} \right) + 1} \right\} \cr & = 1 \cr} $$
$$\eqalign{ & {\text{LH limit}} = \mathop {\lim }\limits_{x \to 0 - 0} g\left\{ {f\left( x \right)} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} g\left\{ {f\left( {0 - h} \right)} \right\} \cr & = \mathop {\lim }\limits_{h \to 0} \left\{ {{{\sin }^2}\left( {0 - h} \right) + 1} \right\} \cr & = 1 \cr} $$