Answer
$$\lim _{x \rightarrow1^{+}} \sqrt{\frac{x-1}{x+2}}=0$$
Work Step by Step
Given $$\lim _{x \rightarrow1^{+}} \sqrt{\frac{x-1}{x+2}}$$
The function is defined when coming from the right
of $1$ (it is actually defined on and around 1 on both
sides), so the limit is just the function value.
So, we get
$$\begin{aligned}L&=\lim _{x \rightarrow 1^{+}} \sqrt{\frac{x-1}{x+2}}\\
&=\sqrt{\frac{1-1}{1+2}}\\
&=\sqrt{\frac{0}{3}}=0
\end{aligned}$$
