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