## Thomas' Calculus 13th Edition

$$\lim _{x \rightarrow1^{+}} \sqrt{\frac{x-1}{x+2}}=0$$
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}