# Proof of Limit Laws

1. Jan 1, 2012

### Alfredoz

Dear All,

I need help on proving:

According to Pauls Online Notes,
we let ε > 0. Since [PLAIN]http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs_files/eq0075M.gif, [Broken] there's a http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs_files/eq0076M.gif such that

I understand that l g(x) - L l < ε, but how do we derive l g(x) - L l < l L l / 2 ?

Last edited by a moderator: May 5, 2017
2. Jan 1, 2012

### SammyS

Staff Emeritus
What does it mean (in terms of δ and ε) for $\displaystyle\lim_{x\,\to\,a}\ g(x)=L\ ?$
...

In particular, if you let ε = |L|/2, then you know that there is some number, call it δ1, such that whenever 0 < | x - a | < δ1,
then | g(x) - L | < |L|/2 .

Last edited by a moderator: May 5, 2017
3. Jan 1, 2012

### Alfredoz

Hi SammyS, thanks for reply. But how do you get | L |/2 ?

4. Jan 1, 2012

### SammyS

Staff Emeritus
Often, the way you come up with such quantities is to "play around" with the δ and the ε expressions, (often working backwards, so to speak, that is -- from the ε to the δ) to find a relationship between δ and ε.

So, off hand, I don't know why he chose |L|/2 , but I haven't gone through his whole argument.

It's just that he's using the |L|/2 for ε, so we know δ1 exists from the definition of the limit and knowing that $\displaystyle\lim_{x\,\to\,a}\,g(x)=L\,.$

5. Jan 1, 2012

### Alfredoz

http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx( [Broken] Proof of 4)

Last edited by a moderator: May 5, 2017
6. Jan 1, 2012

### SammyS

Staff Emeritus
Yes, after looking at this (I had already found it.), maybe I should ask you "What is the role of |L|/2 in the proof?"

BTW, there is nothing magic about |L|/2 itself. All that's needed is something less than |L|.

He uses |L|/2 to get δ1. He uses δ1 to get a bound on 1/|g(x)| , because when you consider what needs to be proved, you realize that you must have
$\displaystyle\left|\frac{1}{g(x)}-\frac{1}{L}\right|<\varepsilon$​
A little algebra shows that
$\displaystyle\left|\frac{1}{g(x)}-\frac{1}{L}\right|=\frac{1}{|L|}\frac{1}{|g(x)|} \left|g(x)-L\right|$​
Looking at details in the proof, we see that δ1 is used so that there is an upper bound on $\displaystyle\frac{1}{|g(x)|}$. δ2 is used so that there is a bound on |g(x)-L| .

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7. Jan 2, 2012

### Alfredoz

May I know why must l g(x) - L l < l L l ?

Last edited: Jan 2, 2012
8. Jan 2, 2012

### SammyS

Staff Emeritus
Actually, if you look at the Notes, $\displaystyle \left|g(x)-L\right|<\frac{L^2}{2}\varepsilon\,.$

This was chosen so that
$\displaystyle \frac{1}{|L|}\frac{1}{|g(x)|} \left|g(x)-L\right|<\frac{1}{|L|}\frac{2}{|L|}\frac{L^2}{2} \varepsilon\,,$​
whenever
$0<|x-a|<\min(\delta_1,\delta_2)\,.$​

9. Jan 2, 2012

### Alfredoz

And how do we know that: l g(x) - L l < (L x L)ε/2 ?

10. Jan 2, 2012

### HallsofIvy

You seem to be completely missing the point! If $\lim_{x\to a} g(x)= L$, then, for any number $\epsilon> 0$, there exist $\delta$ such that if $|x- a|< \delta$ then $|g(x)- L|< \epsilon$. Because $\epsilon$ can be any, we can, if we wish, take it to be L, or L/2, or $(L^2/2)\epsilon$. The only thing that changes is how small $\delta$ must. And if we have $\delta_1$ such that if $|x- a|<\delta_1$ then $|g(x)-L|< \epsilon$ and, say, $\delta_2$ such that if $|x- a|< \delta_2$ then $|g(x)- L|<(L^2/2)\epsilon$, then taking $\delta$ to be the smaller of the two, if $|x- a|<\delta$ then both $|x- a|< \delta_1$ and $|x- a|< \delta_2$ are true and so both $g(x)- L|< \epsilon$ and $|g(x)- L|< (L^2/2)\epsilon$ are true.

11. Jan 2, 2012

### Alfredoz

Ah yes, I get what you mean:) One needs to vary parameters according to the situation:) Thank you v. much!