Answer
$\lim_{x\to1^+}\sqrt\frac{x-1}{x+2}=0$
Work Step by Step
$$\lim_{x\to1^+}\sqrt\frac{x-1}{x+2}$$
To find one-side limit algebraically, we still apply the limit laws like for the two-side limits as normal.
$$\lim_{x\to1^+}\sqrt\frac{x-1}{x+2}=\sqrt{\frac{1-1}{1+2}}=\sqrt\frac{0}{3}=\sqrt0=0$$
Therefore, $\lim_{x\to1^+}\sqrt\frac{x-1}{x+2}=0$
