Answer
$$\frac{8}{3}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {2, - 1,2} \right)} \frac{{x{z^2}}}{{\sqrt {{x^2} + {y^2} + {z^2}} }} \cr
& {\text{Using the limit laws }} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {2, - 1,2} \right)} x{z^2}}}{{\mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {2, - 1,2} \right)} \sqrt {{x^2} + {y^2} + {z^2}} }} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {2, - 1,2} \right)} x{z^2}}}{{\sqrt {\mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {2, - 1,2} \right)} {x^2} + \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {2, - 1,2} \right)} {y^2} + \mathop {\lim }\limits_{\left( {x,y,z} \right) \to \left( {2, - 1,2} \right)} {z^2}} }} \cr
& {\text{Evaluating the limit}} \cr
& = \frac{{\left( 2 \right){{\left( 2 \right)}^2}}}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2}} }} \cr
& = \frac{8}{{\sqrt {4 + 1 + 4} }} \cr
& = \frac{8}{3} \cr} $$
