$\lim_ {x\rightarrow−2} \frac{1}{x + 1} = −1$ Been repeatedly stumped by this limits, epsilon-delta question: 
$\lim_
{x\rightarrow−2}
\frac{1}{x + 1} = −1$.
The answer in the back of Anton Transcendentals is: Min$(\frac{1}{2}, \frac{\epsilon}{2})$
I have tried manipulating it every which way, but cannot get my numbers to come out right, any help greatly appreciated.
$f(x)=|\dfrac{1}{x+1}-(-1)|=|\dfrac{x-(-2)}{x+1}|.$

For $|x-(-2)| &lt;1/2,$ we have

$-1/2 &lt; x+2&lt;1/2$, $-3/2 &lt;x+1&lt;-1/2$, i.e.

$|x+1| &gt;1/2$.

Let $\epsilon &gt;0$ be given.

Choose $\delta =\min(1/2,\epsilon/2)$.

Then $|x-(-2)|&lt;\delta$ implies 

$f(x)=\dfrac{|x+2|}{|x+1|} &lt; \dfrac{\delta}{1/2}= 2\delta &lt;\epsilon.$
If we set $f(x)=\frac{1}{x+1}$, then
$$f(-2)=\frac{1}{-2+1}=-1\in\mathbb{R}$$

Hence,
$$\lim_{x\to-2}f(x)=f(-2)=-1$$

No need to manipulate the numbers .