Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^4} - 16{y^4}}}{{{x^2} + 4{y^2}}} \cr
& {\text{Evaluating the limit by direct substitution, we obtain }}\frac{0}{0} \cr
& {\text{Factoring the difference of two squares for the numerator}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\left( {{x^2} - 4{y^2}} \right)\left( {{x^2} + 4{y^2}} \right)}}{{{x^2} + 4{y^2}}} \cr
& {\text{cancel the common factor}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \left( {{x^2} - 4{y^2}} \right) \cr
& {\text{then}}{\text{, substituting 0 for }}x{\text{ and }}y \cr
& = {\left( 0 \right)^2} - 4{\left( 0 \right)^2} \cr
& = 0 \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{{x^4} - 16{y^4}}}{{{x^2} + 4{y^2}}} = 0 \cr} $$
