Answer
$$\frac{1}{4}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 4} - 2}}{{{x^2}}} \cr
& {\text{Evaluatig}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 4} - 2}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{0^2} + 4} - 2}}{{{{\left( 0 \right)}^2}}} = \frac{0}{0} \cr
& {\text{rationalizing}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 4} - 2}}{{{x^2}}} \times \frac{{\sqrt {{x^2} + 4} + 2}}{{\sqrt {{x^2} + 4} + 2}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {{x^2} + 4} - 2} \right)\left( {\sqrt {{x^2} + 4} + 2} \right)}}{{{x^2}\left( {\sqrt {{x^2} + 4} + 2} \right)}} \cr
& {\text{Use the special product }}\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {\sqrt {{x^2} + 4} } \right)}^2} - {{\left( 2 \right)}^2}}}{{{x^2}\left( {\sqrt {{x^2} + 4} + 2} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2} + 4 - 4}}{{{x^2}\left( {\sqrt {{x^2} + 4} + 2} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{{{x^2}\left( {\sqrt {{x^2} + 4} + 2} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {{x^2} + 4} + 2}} \cr
& {\text{Evaluate}} \cr
& = \frac{1}{{\sqrt {{{\left( 0 \right)}^2} + 4} + 2}} = \frac{1}{4} \cr
& {\text{Then}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 4} - 2}}{{{x^2}}} = \frac{1}{4} \cr} $$
