Answer
$$4a\sqrt a $$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to a} \frac{{{x^2} - {a^2}}}{{\sqrt x - \sqrt a }},\,\,\,\,a > 0 \cr
& {\text{Try to evaluate the limit}} \cr
& = \mathop {\lim }\limits_{x \to a} \frac{{{a^2} - {a^2}}}{{\sqrt a - \sqrt a }} = \frac{0}{0} \cr
& {\text{Rationalizing the denominator}} \cr
& = \mathop {\lim }\limits_{x \to a} \frac{{{x^2} - {a^2}}}{{\sqrt x - \sqrt a }} \times \frac{{\sqrt x + \sqrt a }}{{\sqrt x + \sqrt a }} \cr
& = \mathop {\lim }\limits_{x \to a} \frac{{\left( {{x^2} - {a^2}} \right)\left( {\sqrt x + \sqrt a } \right)}}{{{{\left( {\sqrt x } \right)}^2} - {{\left( {\sqrt a } \right)}^2}}} \cr
& {\text{Simplifying}} \cr
& = \mathop {\lim }\limits_{x \to a} \frac{{\left( {x + a} \right)\left( {x - a} \right)\left( {\sqrt x + \sqrt a } \right)}}{{x - a}} \cr
& = \mathop {\lim }\limits_{x \to a} \left( {x + a} \right)\left( {\sqrt x + \sqrt a } \right) \cr
& {\text{Evaluating the limit}} \cr
& = \left( {a + a} \right)\left( {\sqrt a + \sqrt a } \right) \cr
& = 4a\sqrt a \cr} $$
