# Strange Limit Proof

• December 15th 2008, 03:42 PM
billa
Strange Limit Proof
I thought I understood how to show that a function has a limit using the limit definition, but my book is doing a problem in a way I can't follow. The question is "prove that the limit of x^2 as x approaches 2 is 4"
The book starts with the epsilon equation
$
\left| {x^2 - 4} \right| < \varepsilon
$

and they show that
$
\sqrt {4 - \varepsilon } < x < \sqrt {4 + \varepsilon }
$

which I can understand (though somehow they assume epsilon is less than 4)

Then they say that they want to find delta. They show that delta is
$
Min\{ 2 - \sqrt {4 - \varepsilon } ,\sqrt {4 + \varepsilon } - 2\}
$

Which I am now able to follow although it took awhile to see why this is true. Up to this point I can understand what is going on. This is all the "scratch work" if I am correct. Now they have to actually prove the limit and they say the following...

"If delta has this or any smaller positive value, the inequality
0<abs(x-2)<delta
will place x between
$
\sqrt {4 - \varepsilon } \_and\_\sqrt {4 + \varepsilon }
$

This completes the proof."

I don't understand how this completes the proof. I can prove the limit a different way but it is disheartening that I can't follow this. Can anyone explain how their work proves the limit. Thanks.
• December 15th 2008, 04:20 PM
Mathstud28
Quote:

Originally Posted by billa
I thought I understood how to show that a function has a limit using the limit definition, but my book is doing a problem in a way I can't follow. The question is "prove that the limit of x^2 as x approaches 2 is 4"
The book starts with the epsilon equation
$
\left| {x^2 - 4} \right| < \varepsilon
$

and they show that
$
\sqrt {4 - \varepsilon } < x < \sqrt {4 + \varepsilon }
$

which I can understand (though somehow they assume epsilon is less than 4)

Then they say that they want to find delta. They show that delta is
$
Min\{ 2 - \sqrt {4 - \varepsilon } ,\sqrt {4 + \varepsilon } - 2\}
$

Which I am now able to follow although it took awhile to see why this is true. Up to this point I can understand what is going on. This is all the "scratch work" if I am correct. Now they have to actually prove the limit and they say the following...

"If delta has this or any smaller positive value, the inequality
0<abs(x-2)<delta
will place x between
$
\sqrt {4 - \varepsilon } \_and\_\sqrt {4 + \varepsilon }
$

This completes the proof."

I don't understand how this completes the proof. I can prove the limit a different way but it is disheartening that I can't follow this. Can anyone explain how their work proves the limit. Thanks.

Assuming I understand you correctly proving that works because

$|x^2-4|<\varepsilon\implies -\varepsilon

Etc. Is that what you are asking?
• December 15th 2008, 04:28 PM
billa
umm no lol, but ty still. What i don't understand is given the books definition of delta as the minimum of two expressions, why is this statement true

0<abs(x-2)<delta
will place x between
http://www.mathhelpforum.com/math-he...b747d804-1.gif

and if that statement is true, why does it prove the limit?
• December 15th 2008, 05:25 PM
A common definition for the limit is $\lim_{x\to a}f(x) = b \textrm{ iff } \forall \epsilon>0, \exists \delta>0 \textrm{ such that } |x-a|<\delta \implies |f(x)-b|<\epsilon
$

This is a definition of $\delta$ that exists and is >0 whenever $\epsilon>0$. So all we need to show now is that this definition of $\delta$ satisfies $|x-2|<\delta \implies |x^2-4|<\epsilon$ You can usually do this by reading how you got the definition for $\delta$ backwards, although this does not constitute a formal proof. To do it formally,

Assume x>2 so we can replace |x-2| with x-2.
$0
$2
Find the minimum (Warning: the following is not standard notation)
$4-\sqrt{4-\epsilon}?\sqrt{4+\epsilon}$
both of these are now >0 so we can safely square
$16-8\sqrt{4-\epsilon}+4-\epsilon?4+\epsilon$
$16-2\epsilon?8\sqrt{4-\epsilon}$
Still both >0
$256-64\epsilon+4\epsilon^2?64(4-\epsilon)$
$256-64\epsilon+4\epsilon^2?256-64\epsilon$
so ?=>
and so $\sqrt{4+\epsilon}$ is the minimum, so
$2

Repeating this process assuming x<2 produces $\sqrt{4-\epsilon}
so we can say that $\sqrt{4-\epsilon}

It is very easy to show that this is equivalent to $|x^2-4|<\epsilon$

Quote:

though somehow they assume epsilon is less than 4
When dealing with limits you only need to worry about the small values of epsilon, as how the function behaves when x = 9 doesn't affect the limit as $x \to 1$