# Formal definition of limits and proofs using them.

• Sep 9th 2007, 05:48 PM
ten_twentyfour
Formal definition of limits and proofs using them.
I have a couple of problems due soon that I need help comprehending.

Firstly, I need to prove the Squeeze Theorem using the formal (Epsilon-Delta) definition of limits. I've read the actual proof from Wikipedia, but I just can't wrap my head around it, it feels like they're skipping some sort've explination in certain parts.

For example, on the Wiki page (Squeeze theorem - Wikipedia, the free encyclopedia) it gets to a point where they give http://upload.wikimedia.org/math/3/d...0b53c4e829.png, then jump down and say that this proves http://upload.wikimedia.org/math/c/9...9981ab2106.png, but I don't see how they got from one step to the other.

Then, while they prove the general case, they have http://upload.wikimedia.org/math/0/d...2a89301554.png and end up with http://upload.wikimedia.org/math/f/7...42978c53f5.png. I understand the two steps they show in between, but I don't see how using those two steps results in the latter equation.

Secondly, I need to prove that: lim x->infinity f(x) is equal to lim x->0+ f(1/x), not intuitively, but using:

The definition of lim x-> infinity f(x) = L is: for all epsilon > 0 there exists an N > 0 such that x > N -> |f(x)-L| < epsilon.

Preemptive thanks for the help!
• Sep 9th 2007, 07:03 PM
tukeywilliams
If $\displaystyle 0 < |x-a| < \delta$ then $\displaystyle |f(x)| < \epsilon$ which implies that $\displaystyle \lim_{x \to a} f(x) = 0$ (using definition of limit).
• Sep 9th 2007, 07:08 PM
ten_twentyfour
Oh duh, I totally get that part now. It's kinda hard to follow logic when they're bombarding you with different equations and function representations.
• Sep 9th 2007, 07:12 PM
ThePerfectHacker
$\displaystyle \lim_{x\to a}f(x) = L$ and $\displaystyle \lim_{x\to a}g(x) = L$ means:
$\displaystyle 0<|x-a|<\delta_1 \implies |f(x)-L|<\epsilon$.
$\displaystyle 0<|x-a|<\delta_2 \implies |g(x)-L|<\epsilon$
Choose $\displaystyle \delta = \min \{ \delta_1,\delta_2 \}$.
This will imply that,
$\displaystyle 0<|x-a|<\delta \implies |f(x)-L|<\epsilon \mbox{ and }|g(x)-L|<\epsilon$.
Now argue that since,
$\displaystyle g(x)\leq h(x)\leq f(x)$
We must have that,
$\displaystyle |h(x) - L|<\epsilon \mbox{ for }0<|x-a|<\delta$.
• Sep 9th 2007, 07:42 PM
ten_twentyfour
I've officially solved the first question, I just had to sit there and stare at the proof until it finally clicked. Now to work on the other one.