Answer
$$\lim _{(x, y) \rightarrow(2,1)} \frac{x-y-1}{\sqrt{x-y}-1}=2 $$
Work Step by Step
Given$$\lim _{(x, y) \rightarrow(2,1)} \frac{x-y-1}{\sqrt{x-y}-1} $$
So, by multiplying the numerator and denominator by the conjugate of the denominator, we get
\begin{align}
L&=\lim _{(x, y) \rightarrow(2,1)} \frac{x-y-1}{\sqrt{x-y}-1} \\
&=\lim _{(x, y) \rightarrow(2,1)} \frac{x-y-1}{\sqrt{x-y}-1} \cdot \frac{\sqrt{x-y}+1}{\sqrt{x-y}+1} \\
&=\lim _{(x, y) \rightarrow(2,1)} \frac{(x-y-1)(\sqrt{x-y}+1)}{(\sqrt{x-y})^2-1^2} \\
&=\lim _{(x, y) \rightarrow(2,1)} \frac{(x-y-1)(\sqrt{x-y}+1)}{(x-y-1)} \\ &=\lim _{(x, y) \rightarrow(2,1)}(\sqrt{x-y}+1)\\
&= \sqrt{2-1}+1\\
&=2
\end{align}