Question

(23 Points) Prove that the limit is correct:

lim x approaches -2 of 1/(x+1) = -1

Answer 1

The statement that

`\displaystyle\lim_(x\to-2)\frac1{x+1}=-1`

is equivalent to saying that, for any
`\varepsilon>0`, there exists
`\delta>0` such that if

`0<|x+2|<\delta`

then it's guaranteed that

`\left|\frac1{x+1}+1\right|<\varepsilon`

We want to pick
`\delta` to make this guarantee.

We have

`\left|\frac1{x+1}+1\right|=\left|(x+2)/(x+1)\right|=(|x+2|)/(|x+1|)`

If we assume
`\delta\le\frac12`, then

`|x+2|<\frac12\implies-\frac12<x+2<\frac12\implies-\frac32<x+1<-\frac12\implies\frac12<|x+1|<\frac32`

The lower bound on
`|x+1|` is what's important here, because its gives us an upper bound on the reciprocal:

`\frac23<\frac1<\frac12`

Then in the expected
`\varepsilon`-inequality, we have

`(|x+2|)/(|x+1|)<\frac2<\varepsilon\implies|x+2|<2\varepsilon`

Comparing this to our chosen
`\delta`-inequality, this suggests we should pick
`\delta` to be the smaller of
`\frac12` and
`2\varepsilon`, or
`\delta=\min\left\{\frac12,2\varepsilon\right\}`.

# # #

Now for the proof itself. Let
`\varepsilon>0` be given, and assume
`|x+2|<\min\left\{\frac12,2\varepsilon\right\}`.

• If
  `\frac12<2\varepsilon`, then
  `\frac14<\varepsilon` and

`|x+2|<\frac12\implies(|x+2|)/(|x+1|)=\left|\frac1{x+1}+1\right|<\frac14<\varepsilon`

• If
  `\frac12>2\varepsilon`, then

`|x+2|<2\varepsilon\implies\fracx+22<\varepsilon`

and

`(|x+2|)/(|x+1|)=\left|\frac1{x+1}+1\right|<\fracx+22<\varepsilon`

This completes the proof.