Answer
$=\dfrac{5}{2}$
Work Step by Step
Solve the limit $\lim\limits_{(x,y) \to (0,0)}\dfrac{3x^2-y^2+5}{x^2+y^2+2}$.
Thus, we have
$\lim\limits_{(x,y) \to (0,0)}\dfrac{3x^2-y^2+5}{x^2+y^2+2}=\dfrac{3(0)^2-(0)^2+5}{(0)^2+(0)^2+2}=\dfrac{5}{2}$
