Epsilon-delta proof: lim (x^2 + 3) = 4 as x -> 1

[raaky217]

I just cannot figure this out. Here is a problem I worked on and could not figure out. This is the explaination given by professor but its still confusing. Can any one break this down into lamens terms?

lim (x^2 + 3) = 4
x->1

0 < |x-1| < d, then |f(x) - 4| < e.

|f(x) - 4|< e iff |(x^2 + 3)- 4|< e

|x^2 - 1|< e

|(x-1)(x+1)< e

|x-1| |x+1| < e

I understand this up to this point. Professor goes on to write this:

|x-1| |x+1| < |x-1|(3) < e ?????????? Where did the 3 come from?

|x-1| (3) < e

|x-1| < e/3

Choose d=min {1, e/3}

Please help if you can.[/code]

 

[pka]

If |x-1|<1 then |x+1|=|x-1+2|<|x-1|+|2|<3.
|x-1||x+1|<(e/3)(3)=e

 

[soroban]

Hello, raaky217!

This is the explaination given by professor.
Can anyone break this down into layman's terms?

\(\displaystyle \lim_{x\to1}(x^2\,+\,3) \:= \:4\)

\(\displaystyle 0\:<\:|x\,-\,1|\:<\:\delta\), then \(\displaystyle |f(x)\,-\,4|\:<\:\epsilon\)

\(\displaystyle |f(x)\,-\,4|\:<\:\epsilon\;\longleftrightarrow\;|(x^2\,+\,3)\,-\,4|\:<\:\epsilon\)

\(\displaystyle |x^2\,-\,1|\:< \L\epsilon\)

\(\displaystyle |(x\,-\,1)(x\,+\,1) \:< \:\epsilon\)

\(\displaystyle |x\,-\,1|\cdot|x\,+\,1| \:< \:\epsilon\)

I understand this up to this point.
Professor goes on to write this:

\(\displaystyle |x\,-\,1|\cdot|x\,+\,1|\:<\:|x\,-\,1|(3) \:< \:\epsilon\;\;\) . . . Where did the 3 come from?

\(\displaystyle |x\,-\,1|(3) \:< \:\epsilon\)

\(\displaystyle |x\,-\,1|\: < \:\frac{\epsilon}{3}\)

Choose: \(\displaystyle \,\delta \:=\:min\left(1,\,\frac{\epsilon}{3}\right)\)

It was explained to me in baby-talk like this . . .

Since \(\displaystyle x\to1,\;x\) is "very close" to 1,
. . then \(\displaystyle \,|x\,+\,1|\) is "very close" to 2.
Hence: \(\displaystyle \,|x\,+\,1|\:<\:3\)

Multiply both sides by \(\displaystyle |x\,-\,1|\)
. . and we have: \(\displaystyle \,|x\,-\,1|\cdot|x\,+\,1| \:< |x\,-\,1|(3)\)

Got it?