Answer
$$\lim\limits_{(x,y) \to (2,1)} = \frac{2}{7}$$
Work Step by Step
$xy$, $x^2$ and $3y^2$ are continuous everywhere (polynomial functions).
$\frac{4-xy}{x^2 + 3y^2}$ is continuous on its domain (rational function): $x^2 + 3y^2 \ne 0$
In this case: $(2)^2 + 3(1)^2 = 7 \ne 0$. Thus, this function is continuous at (2,1).
Therefore, we can substitute the values of x and y directly into the fuction.
$$\lim\limits_{(x,y) \to (2,1)} \Bigg(\frac{4 - xy}{x^2 + 3y^2}\Bigg) = \Bigg(\frac{4 - (2)(1)}{(2)^2 + 3(1)^2}\Bigg)$$ $$= \frac{2}{7}$$
