Prove the stated limit is correct...
$$\lim_{x → -2}\frac{1}{(x + 1)} = -1 $$
So far I have managed to reduce it to
$$|\frac{1}{(x + 1)}||x+2| < ε \text{ if } 0<|x+2|<δ$$
and then I would usually set delta less or equal to one and go from there,
$$|x+2|<1$$
and try rearrange it to $$ \text{ stuff } < \frac{1}{(x + 1)}< \text{ stuff }$$
Im not really sure how to do this part. I've only been learning epsilon-delta from this morning so I would appreciate a thorough explanation.
You want to conclude that $$\left| \dfrac 1 {x+1} \right| \left| x+2 \right|<\varepsilon.$$ So you need to figure out what to do with $\left| \dfrac 1 {x+1} \right|.$ The fact that $x$ is between $-2\pm\delta$ enables us to say something about $\left| \dfrac 1 {x+1} \right|$ provided we know something about $\delta.$ If $\delta$ is small the two numbers $-2\pm\delta$ are negative, and we have \begin{align} & -2-\delta < x < -2+\delta \\[10pt] & -1-\delta < x+1 < -1 + \delta \\[10pt] & \frac 1 {-1-\delta} > \frac 1 {x+1} > \frac 1 {-1+\delta} \\[10pt] & \frac 1 {1+\delta} < \left| \frac 1 {x+1}\right| < \frac 1 {1-\delta} \\ & \quad \text{Here we relied on the fact that all three numbers were negative,} \\ & \quad \text{so taking absolute values amounts to multiplication by }-1. \\[10pt] \text{So } & \left| \frac 1 {x+1} \right| \le 2 \text{ if } \delta\le \frac 1 2. \\[10pt] \text{And then } & \left| \frac 1 {x+1} \right| |x+2| < 2\delta \text{ if } |x+2|<\delta. \\[10pt] \text{We want } & \left| \frac 1 {x+1} \right| |x+2| < \varepsilon. \\[12pt] \text{So let } & \delta = \begin{cases} 1/2, \\ \varepsilon/2, \end{cases} \text{ whichever is less.} & \end{align}