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