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
 A: The function 
$$f(x)=\sqrt{\frac{x+2}{x+1}}$$ 
is continuous at the point in question, so you have that
$$\lim_{x\rightarrow-0.5^-}\sqrt{\frac{x+2}{x+1}}=\lim_{x\rightarrow -0.5^+}\sqrt{\frac{x+2}{x+1}}=\sqrt{\frac{-.5+2}{-.5+1}}=\sqrt{\frac{1.5}{.5}}\\
=\sqrt{3}
$$
Since for a function continuous at a point $a$ you have
$$
\lim_{x\rightarrow a^-}f(x)=\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a}f(x)=f(a)
$$
A: There exist functions which have a 'left hand' limit different from the 'right hand' limit. Or perhaps the limit exists on one side at a particular number but fails to exist on the other. Consider the function
\begin{equation}
f(x)=\dfrac{2x-1}{\vert 2x-1\vert}
\end{equation}
\begin{equation}
\lim_{x\to0.5^-}f(x)=-1
\end{equation}
yet
\begin{equation}
\lim_{x\to0.5^+}f(x)=+1
\end{equation}
So
\begin{equation}
\lim_{x\to0.5}f(x)\text{ does not exist.}
\end{equation}
Notice that $f(x)$ is undefined at $\tfrac{1}{2}$ so one cannot substitute into the function to find the limit.
A: If the limit itself exists, than the left- and right-sided limits coincide. The easiest approach is then to simply compute
$$\lim_{x \to \frac{1}{2}} \sqrt{\frac{x+2}{x+1}}$$
A: *

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


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:
$$\sqrt{\lim_{x \to 0.5}\frac{x+2}{x+1}} = \sqrt{\lim_{x \to 0.5^{-1}}\frac{x+2}{x+1}} \ \text{by continuity of} \ \frac{x+2}{x+1}$$
$$\sqrt{\lim_{x \to 0.5}\frac{x+2}{x+1}} = \lim_{x \to 0.5}\sqrt{\frac{x+2}{x+1}} \ \text{by continuity of} \ \sqrt{x}$$
$$\lim_{x \to 0.5^{-1}}\sqrt{\frac{x+2}{x+1}} = \lim_{x \to 0.5}\sqrt{\frac{x+2}{x+1}} \ \text{by continuity of} \ \sqrt{x}$$
