How do you find one-sided limits *algebraically*? Find $$\lim_{x\to\  -0.5^-}\sqrt{\frac{x+2}{x+1}}$$ 
Sorry, I have no idea where to start. I know how to find regular limits algebraically, but not one-sided.
Thanks

*

*$$\lim_{x \to -0.5} f(x) = \lim_{x \to -0.5^{-1}} f(x)$$


if $f$ is continuous at $x=-0.5$
Is $$\sqrt{\frac{x+2}{x+1}}$$ continuous at $x=-0.5$?
What we need to check is if $$\lim_{x \to -0.5} \sqrt{x} = \lim_{x \to -0.5^{-1}} \sqrt{x}$$
and if $$\lim_{x \to 0-.5} \frac{x+2}{x+1} = \lim_{x \to -0.5^{-1}} \frac{x+2}{x+1}$$
It turns out both are true because


*

*$\sqrt{x}$ is continuous at all points $x > 0$ and $\frac{x+2}{x+1}|_{x = -0.5} > 0$.

*$\frac{x+2}{x+1}$ is continuous at all points $x \ne -1$
So this what we have:
$$\sqrt{\lim_{x \to 0.5}\frac{x+2}{x+1}} = \sqrt{\lim_{x \to 0.5^{-1}}\frac{x+2}{x+1}} \ \text{by continuity of} \ \frac{x+2}{x+1}$$
$$\sqrt{\lim_{x \to 0.5}\frac{x+2}{x+1}} = \lim_{x \to 0.5}\sqrt{\frac{x+2}{x+1}} \ \text{by continuity of} \ \sqrt{x}$$
$$\lim_{x \to 0.5^{-1}}\sqrt{\frac{x+2}{x+1}} = \lim_{x \to 0.5}\sqrt{\frac{x+2}{x+1}} \ \text{by continuity of} \ \sqrt{x}$$
