Disproving Lim x->2 (x^2 -1) = 3

[Jaskaran]

From the limit laws, we know that lim x->2 (x^2 - 1) = 3. Finding a number ? such that if 0<|x-2|<? then |(x^2 - 1) - 3| < 0.2

-0.2 <|(x^2 - 1) - 3| < 0.2
14/5 <(x^2 -1)<16/5
19/5 < x^2 < 21/5
1.949 < x < 2.049
1.949 - 2 < x - 2 < 2.049 - 2
-.0506 < x < 0.0493

Then 0<|x-2|<?

? ~ 0.0493 or any smaller positive number.

BUT this does NOT prove that Lim x->2 (x^2 -1) = 3

Why?

 

[DrMike]

What is the definition of a limit?

\(\displaystyle \forall\epsilon>0,\,\exists\delta...\)

But the fake "proof" only checks \(\displaystyle \epsilon=0.2\).

Has it proven that for all epsilon, there exists a delta such that bla bla bla?

 

[galactus]

\(\displaystyle |(x^{2}-1)-3|=|x^{2}-4|=|x+2||x-2|\)

If we restrict \(\displaystyle {\delta}\) so that \(\displaystyle {\delta}\leq 1\), then

\(\displaystyle |x-2|<1\)

\(\displaystyle -1<x-2<1\)

\(\displaystyle 1<x<3\)

\(\displaystyle 3<x+2<5\)

which implies:

\(\displaystyle |x+2|<5\)

\(\displaystyle |x+2||x-2|\leq 5|x-2|\)

\(\displaystyle \therefore \;\ |(x^{2}-1)-3|<{\epsilon}\)

if \(\displaystyle 5|x-2|<{\epsilon}\)

or if \(\displaystyle |x-2|<\frac{\epsilon}{5}\)

So, \(\displaystyle {\delta}=min(\frac{\epsilon}{5},1)\)