The statement that
[tex]\displaystyle\lim_{x\to-2}\frac1{x+1}=-1[/tex]
is equivalent to saying that, for any [tex]\varepsilon>0[/tex], there exists [tex]\delta>0[/tex] such that if
[tex]0<|x+2|<\delta[/tex]
then it's guaranteed that
[tex]\left|\dfrac1{x+1}+1\right|<\varepsilon[/tex]
We want to pick [tex]\delta[/tex] to make this guarantee.
We have
[tex]\left|\dfrac1{x+1}+1\right|=\left|\dfrac{x+2}{x+1}\right|=\dfrac{|x+2|}{|x+1|}[/tex]
If we assume [tex]\delta\le\dfrac12[/tex], then
[tex]|x+2|<\dfrac12\implies-\dfrac12<x+2<\dfrac12\implies-\dfrac32<x+1<-\dfrac12\implies\dfrac12<|x+1|<\dfrac32[/tex]
The lower bound on [tex]|x+1|[/tex] is what's important here, because its gives us an upper bound on the reciprocal:
[tex]\dfrac23<\dfrac1{|x+1|}<\dfrac12[/tex]
Then in the expected [tex]\varepsilon[/tex]-inequality, we have
[tex]\dfrac{|x+2|}{|x+1|}<\dfrac{|x+2|}2<\varepsilon\implies|x+2|<2\varepsilon[/tex]
Comparing this to our chosen [tex]\delta[/tex]-inequality, this suggests we should pick [tex]\delta[/tex] to be the smaller of [tex]\dfrac12[/tex] and [tex]2\varepsilon[/tex], or [tex]\delta=\min\left\{\dfrac12,2\varepsilon\right\}[/tex].
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Now for the proof itself. Let [tex]\varepsilon>0[/tex] be given, and assume [tex]|x+2|<\min\left\{\dfrac12,2\varepsilon\right\}[/tex].
[tex]|x+2|<\dfrac12\implies\dfrac{|x+2|}{|x+1|}=\left|\dfrac1{x+1}+1\right|<\dfrac14<\varepsilon[/tex]
[tex]|x+2|<2\varepsilon\implies\dfrac{|x+2|}2<\varepsilon[/tex]
and
[tex]\dfrac{|x+2|}{|x+1|}=\left|\dfrac1{x+1}+1\right|<\dfrac{|x+2|}2<\varepsilon[/tex]
This completes the proof.
