# Homework Help: Another attempt to understand min/max value of lambda

1. Jan 17, 2009

### transgalactic

i read and translated this "article" from a book that shows a new concept to me.
infinity small and infinitely large function.

can you explain to me using those terms

http://img262.imageshack.us/img262/1610/97335712sc3.gif [Broken]

why in certain case we use maximum as the value for lambda and why
in another we use minimum as the value for lambda
like in here:
http://img132.imageshack.us/img132/315/67603225wa1.th.gif [Broken]

Last edited by a moderator: May 3, 2017
2. Jan 17, 2009

### Staff: Mentor

The text that you show has nothing to do with infinitely large or infinitesimally small anything. Also, this Greek letter $\delta$, is lower-case delta, not lambda.

All that the author is doing is showing that
$$\lim_{x \rightarrow 5} x^2 = 25$$

To do that, he has to show that for any given epsilon > 0, there is a number delta > 0 so that |x^2 - 25| < epsilon when |x - 5| < delta.

The author factored |x^2 - 25| into |x - 5| |x + 5|. The trouble is, you can't just divide both sides of the inequality by |x + 5|, since its value depends on x. If you can nail down delta, which represents how close x is to 5, and say how big delta can be at most, then you can say something about the value of |x + 5|. You then pick delta to be the smaller of {1, epsilon/11}, which forces delta to be 1 at most. Then you know that |x + 5| will be between 9 and 11 (since x is between 4 and 6), so |x - 5| |x + 5| <= 11 |x - 5| < 11 * delta = epsilon.

This is pretty standard stuff for working with limits.

Last edited by a moderator: May 3, 2017
3. Jan 17, 2009

### transgalactic

Last edited by a moderator: May 3, 2017
4. Jan 17, 2009

### Staff: Mentor

I can't tell what you have here. A lot of what's on the page seems to be whited out. How does B play a role here? There's not enough to be able to tell what you're trying to do.

Why don't you type this stuff into your post rather than writing it on paper and then scanning it and uploading an image? It would be easier (and less irritating) for us readers to see what you have all in one place.

5. Jan 17, 2009

### transgalactic

its a bounding of f(x) prove
next next line says
|f(x)|<B so f(x) is bounded

this is very similar to before
but here in b they used max instead of minimum
??

6. Jan 17, 2009

### Staff: Mentor

I don't see anything in your work that says |f(x)| < B. This is exactly my point. I am asking you to put your work in your post so that it's easy and convenient to see, and is all in one place.

7. Jan 18, 2009

### transgalactic

here is the full work:
http://img292.imageshack.us/img292/8506/67756915is4.gif [Broken]

Last edited by a moderator: May 3, 2017
8. Jan 18, 2009

### Staff: Mentor

Why can't you put what you did right in your post?

What do you mean "and there is a final border"?
Your definition of
$$\lim_{x \rightarrow \infty} f(x) = A$$
isn't correct.
It should be
$$\forall \epsilon > 0 \ \exists c \in R \ni x > c \Rightarrow |f(x) - A| < \epsilon$$

In English, this is "for any/each positive epsilon there exists a real number c such that x > c implies that f(x) is within epsilon of A."

Since you can get f(x) within epsilon of A for x > c, that shows that f is bounded on that interval.

For the interval [a, c] you invoked the Weierstrass Law to show that f was bounded on that interval. What does the Weierstrass Law say? I looked in 5 or more books I have and on the internet. There's a Stone-Weierstrass Theorem and a Bolzano-Weierstrass Theorem, and several other references, but it's not clear what you're using here.

As for the max part you want a number B that's larger than the largest value of f(x) on [a, c] and larger than the largest value of f(x) for x > c. That way you can say that B is larger than any value of f(x) for x >= a. (Not for "every x" as you said.)

9. Jan 18, 2009

### transgalactic

i used math type software to type it
i dont know how to turn it into itex code??

10. Jan 18, 2009

### transgalactic

why they use the Weierstrass law only on the 0<x<c case

why we cant say
for x>c
there is M>|f(x)|

(like in the couchy theorim for bounded sequence)
??

11. Jan 18, 2009

### transgalactic

how whould you prove that its also bounded for
a<x<c
??

12. Jan 18, 2009

### Staff: Mentor

I don't know why "they" did this, since I don't have the book you're using. The work you show is in two parts:
1. A bound exists for x > c.
2. A bound exists for a <= x <= c.
The work for the first part uses the fact that
$$\lim_{x \rightarrow \infty} f(x) = A$$
to show that |f(x)| < A + $\epsilon$, when x > c.

The work for the second part is what I'm questioning. You referred to Weierstrass Law, and I think you mean Bolzano-Weierstrass, but that applies to bounded sequences, so I'm not sure how that fits in here.

Regarding Cauchy theorem for bounded sequences, 1) what sequences are you talking about with your function f(x), and 2) if you use such a theorem, include the statement of this theorem so we understand what you're talking about.

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