Answer
The limit is
$$\lim_{(x,y)\to(1,1)}\frac{x}{\sqrt{x+y}}=\frac{1}{\sqrt{2}}.$$
The function is discontinuous on the line $x+y=0$.
Work Step by Step
We will calculate this limit by a simple substitution
$$\lim_{(x,y)\to(1,1)}\frac{x}{\sqrt{x+y}}=\frac{1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}.$$
The expression under the square root must not be negative and the whole denominator must not be zero so we have $x+y\geq0$ and $\sqrt{x+y}\neq0\Rightarrow x+y\neq0.$ This put together gives $x+y>0$. In every ordered pair $(x_0,y_0)$ such that $x_0+y_0>0$ the function is continuous because we can always substitute and get
$$\lim_{(x,y)\to(x_0,y_0)}\frac{x}{\sqrt{x+y}}=\frac{x_0}{\sqrt{x_0+y_0}}.$$
However, when $x_0+y_0$=0, the denominator is zero, and since we cannot divide by zero we have
$$\lim_{(x,y)\to(x_0,y_0)}\frac{x}{\sqrt{x+y}}\neq\frac{x_0}{\sqrt{x_0+y_0}}$$
so the function has a discontinuity on a line $x+y=0$,
