# Thread: Question concerning delta epsilon proof

1. ## Question concerning epsilon-delta proof

$Lim\frac{\1}{x^2}=\frac{\1}{4}$

$x\rightarrow2$

$\frac{\1}{x^2}-\frac{\1}{4}$

which breaks down to :

$\frac{\(x-2)(x+2)}{4x^2}$

But I'm confused how to bound this to get a delta.
(PS: I cant figure out how to put the absolute value signs in here, but I know they are supposed to be there)

Do I break it down to :

$(x-2)*\frac{\(x+2)}{4x^2}=(x-2)*\frac{\1}{4x}+\frac{\1}{2x^2}$

Then what?

2. ## Re: Question concerning epsilon-delta proof

Originally Posted by Jerry99
$Lim\frac{\1}{x^2}=\frac{\1}{4}$
$x\rightarrow2$
$\frac{\1}{x^2}-\frac{\1}{4}$
which breaks down to :
$\frac{\(x-2)(x+2)}{4x^2}$
But I'm confused how to bound this to get a delta.
(PS: I cant figure out how to put the absolute value signs in here, but I know they are supposed to be there)
[TEX]\left|{\frac{\1}{x^2}-\frac{\1}{4}\right|[/TEX] gives $\left|{\frac{\1}{x^2}-\frac{\1}{4}\right|$

$\left|{\frac{\1}{x^2}-\frac{\1}{4}\right|=\frac{|x-2||x+2|}{|4x^2|}$

If $0<\delta<1~\&~0<|x-2|<\delta$ then $|x+2|<5$ and $\left|{\frac{\1}{4x^2}\right|<\frac{1}{4}$

See what you can do with that.

3. ## Re: Question concerning epsilon-delta proof

I get how $\left|x+2\right|<5$

But how do you get the $\left|\frac{\1}{4x^2}\right|<\frac{\1}{4}$

and thanks for the absolute value sign tip

4. ## Re: Question concerning delta epsilon proof

I think I get what is going on now actually. that has to be less than 1/4 basically

thanks for the help

5. ## Re: Question concerning epsilon-delta proof

Originally Posted by Jerry99
I get how $\left|x+2\right|<5$
But how do you get the $\left|\frac{\1}{4x^2}\right|<\frac{\1}{4}$
If $1 then $4<4x^2<36$ or $\frac{1}{36}<\frac{1}{4x^2}<\frac{1}{4}$.

6. ## Re: Question concerning epsilon-delta proof

Ok cool I get it. Thanks for the help

7. ## Re: Question concerning delta epsilon proof

Sigh , well, I thought I did. What am I missing here that allows you to make the statement of $4 < 4x^2<36 =\frac{\1}{36}<\frac{\1}{4x^2}<\frac{\1}{4}$

My min $\delta$from this : 1, $\frac{\epsilon}{3},\frac{\epsilon}{5}$ ??

8. ## Re: Question concerning delta epsilon proof

Originally Posted by Jerry99
Sigh , well, I thought I did. What am I missing here that allows you to make the statement of $4 < 4x^2<36 =\frac{\1}{36}<\frac{\1}{4x^2}<\frac{\1}{4}$
My min $\delta$from this : 1, $\frac{\epsilon}{4},\frac{\epsilon}{5}$ ??
$\begin{array}{l}1 < x < 3\\ 1 < {x^2} < 9\\ 4 < 4{x^2} < 36\\ \dfrac{1}{{36}} < \dfrac{1}{{4{x^2}}} < \dfrac{1}{4} \end{array}$

9. ## Re: Question concerning delta epsilon proof

if 0 < a < b < c, then 0 < a2 < ab < b2 < bc < c2, or in short:

0 < a2 < b2 < c2.

similarly, if 0 < a < b < c then:

0 < 1/c < 1/b < 1/a

the basis for the last inequality is this:

we define for two fractions a/b, c/d: a/b < c/d if and only if ad < bc. note this agrees with our usual notion if b = d = 1.

this means that if a < c, that when we compare:

1/a and 1/c, we have 1(c) > a(1), so 1/a > 1/c.

for example: 4 is bigger than 2, so 1/4 is smaller than 1/2.