Proving Limit Laws - Pauls Online Math Notes

In summary, the conversation is about proving a limit with the use of ε and δ. The person asking for help is trying to understand how to derive a certain inequality in the proof, and the other person explains that the specific value chosen for ε is not important, as long as it is smaller than the limit. The conversation then delves into the details of the proof and how to choose the right values for δ. The summary concludes that varying parameters is necessary in order to prove the limit.
  • #1
Alfredoz
11
0
Dear All,

I need help on proving:
eq0073M.gif



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
eq0077M.gif




I understand that l g(x) - L l < ε, but how do we derive l g(x) - L l < l L l / 2 ?
 
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  • #2
Alfredoz said:
Dear All,

I need help on proving:
eq0073M.gif



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
eq0077M.gif




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

What does it mean (in terms of δ and ε) for [itex]\displaystyle\lim_{x\,\to\,a}\ g(x)=L\ ?[/itex]
...

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 .
 
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  • #3
Hi SammyS, thanks for reply. But how do you get | L |/2 ?
 
  • #4
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 [itex]\displaystyle\lim_{x\,\to\,a}\,g(x)=L\,.[/itex]
 
  • #5
SammyS said:
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 [itex]\displaystyle\lim_{x\,\to\,a}\,g(x)=L\,.[/itex]


Hi, please refer to:
http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx( [Broken] Proof of 4)
 
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  • #6
Alfredoz said:
Hi, please refer to:
http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx( [Broken] Proof of 4)

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
[itex]\displaystyle\left|\frac{1}{g(x)}-\frac{1}{L}\right|<\varepsilon[/itex]​
A little algebra shows that
[itex]\displaystyle\left|\frac{1}{g(x)}-\frac{1}{L}\right|=\frac{1}{|L|}\frac{1}{|g(x)|} \left|g(x)-L\right|[/itex]​
Looking at details in the proof, we see that δ1 is used so that there is an upper bound on [itex]\displaystyle\frac{1}{|g(x)|}[/itex]. δ2 is used so that there is a bound on |g(x)-L| .
 
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  • #7
SammyS said:
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
[itex]\displaystyle\left|\frac{1}{g(x)}-\frac{1}{L}\right|<\varepsilon[/itex]​
A little algebra shows that
[itex]\displaystyle\left|\frac{1}{g(x)}-\frac{1}{L}\right|=\frac{1}{|L|}\frac{1}{|g(x)|} \left|g(x)-L\right|[/itex]​
Looking at details in the proof, we see that δ1 is used so that there is an upper bound on [itex]\displaystyle\frac{1}{|g(x)|}[/itex]. δ2 is used so that there is a bound on |g(x)-L| .

May I know why must l g(x) - L l < l L l ?
 
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  • #8
Alfredoz said:
May I know why must l g(x) - L l < l L l ?
Actually, if you look at the Notes, [itex]\displaystyle \left|g(x)-L\right|<\frac{L^2}{2}\varepsilon\,.[/itex]

This was chosen so that
[itex]\displaystyle \frac{1}{|L|}\frac{1}{|g(x)|} \left|g(x)-L\right|<\frac{1}{|L|}\frac{2}{|L|}\frac{L^2}{2} \varepsilon\,,[/itex]​
whenever
[itex]0<|x-a|<\min(\delta_1,\delta_2)\,.[/itex]​
 
  • #9
SammyS said:
Actually, if you look at the Notes, [itex]\displaystyle \left|g(x)-L\right|<\frac{L^2}{2}\varepsilon\,.[/itex]

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


And how do we know that: l g(x) - L l < (L x L)ε/2 ?
 
  • #10
You seem to be completely missing the point! If [itex]\lim_{x\to a} g(x)= L[/itex], then, for any number [itex]\epsilon> 0[/itex], there exist [itex]\delta[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|g(x)- L|< \epsilon[/itex]. Because [itex]\epsilon[/itex] can be any, we can, if we wish, take it to be L, or L/2, or [itex](L^2/2)\epsilon[/itex]. The only thing that changes is how small [itex]\delta[/itex] must. And if we have [itex]\delta_1[/itex] such that if [itex]|x- a|<\delta_1[/itex] then [itex]|g(x)-L|< \epsilon[/itex] and, say, [itex]\delta_2[/itex] such that if [itex]|x- a|< \delta_2[/itex] then [itex]|g(x)- L|<(L^2/2)\epsilon[/itex], then taking [itex]\delta[/itex] to be the smaller of the two, if [itex]|x- a|<\delta[/itex] then both [itex]|x- a|< \delta_1[/itex] and [itex]|x- a|< \delta_2[/itex] are true and so both [itex]g(x)- L|< \epsilon[/itex] and [itex]|g(x)- L|< (L^2/2)\epsilon[/itex] are true.
 
  • #11
HallsofIvy said:
You seem to be completely missing the point! If [itex]\lim_{x\to a} g(x)= L[/itex], then, for any number [itex]\epsilon> 0[/itex], there exist [itex]\delta[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|g(x)- L|< \epsilon[/itex]. Because [itex]\epsilon[/itex] can be any, we can, if we wish, take it to be L, or L/2, or [itex](L^2/2)\epsilon[/itex]. The only thing that changes is how small [itex]\delta[/itex] must. And if we have [itex]\delta_1[/itex] such that if [itex]|x- a|<\delta_1[/itex] then [itex]|g(x)-L|< \epsilon[/itex] and, say, [itex]\delta_2[/itex] such that if [itex]|x- a|< \delta_2[/itex] then [itex]|g(x)- L|<(L^2/2)\epsilon[/itex], then taking [itex]\delta[/itex] to be the smaller of the two, if [itex]|x- a|<\delta[/itex] then both [itex]|x- a|< \delta_1[/itex] and [itex]|x- a|< \delta_2[/itex] are true and so both [itex]g(x)- L|< \epsilon[/itex] and [itex]|g(x)- L|< (L^2/2)\epsilon[/itex] are true.

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

1. What are limit laws in mathematics?

Limit laws in mathematics refer to a set of rules that govern the behavior of limits in calculus. These rules allow for simplification and manipulation of limits, making it easier to solve complex problems.

2. Why is it important to understand limit laws?

Understanding limit laws is crucial for solving many mathematical problems, especially in calculus. These laws help us to evaluate limits more efficiently and accurately.

3. How can limit laws be proven?

The most common method for proving limit laws is through mathematical induction. This involves proving the base case and then the inductive step for each law.

4. What are some examples of limit laws?

Some examples of limit laws include the sum law, product law, quotient law, and power law. These laws state that the limit of a sum, product, quotient, or power of two functions is equal to the sum, product, quotient, or power of their respective limits.

5. Can limit laws be applied to all types of functions?

Yes, limit laws can be applied to all types of functions, including polynomial, rational, exponential, and trigonometric functions. However, certain special cases may require additional techniques for evaluation.

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