Answer
$ + \infty $
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2}}}{{{{\left( {x - 2} \right)}^2}}} \cr
& {\text{Evaluate the limit by direct substitution}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2}}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{{{\left( 2 \right)}^2}}}{{{{\left( {2 - 2} \right)}^2}}} = \frac{4}{0}{\text{ }} \cr
& {\text{Finding the one sided limits }}\mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2}}}{{{{\left( {x - 2} \right)}^2}}}{\text{ and }}\mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2}}}{{{{\left( {x - 2} \right)}^2}}} \cr
& \mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2}}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{{{\left( {{2^ - }} \right)}^2}}}{{{{\left( {{2^ - } - 2} \right)}^2}}} = \frac{4}{{{{\left( {{0^ - }} \right)}^2}}} = \frac{4}{{{0^ + }}} = + \infty \cr
& and \cr
& \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2}}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{{{\left( {{2^ + }} \right)}^2}}}{{{{\left( {{2^ + } - 2} \right)}^2}}} = \frac{4}{{{{\left( {{0^ + }} \right)}^2}}} = \frac{4}{{{0^ + }}} = + \infty \cr
& \mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2}}}{{{{\left( {x - 2} \right)}^2}}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2}}}{{{{\left( {x - 2} \right)}^2}}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2}}}{{{{\left( {x - 2} \right)}^2}}} = + \infty \cr} $$
