Math Is Fun Forum

Hi gAr and hi Bobbym, indeed it is!

This is using the δ-ϵ (delta-epsilon) definition to prove limits. Then there is the N-ϵ method to prove limits which involve infinity.

So for the δ-ϵ method, it is basically saying the following: if we have lim of x -> a f(x) = L. For all x, if 0 < |x-a| < δ (we choose this δ), then |f(x) - L| < ϵ.

We take ϵ > 0 to be arbitrary and then prove the implication (the then statement). If the proof is successful, then the limit has been proven. The N-ϵ is similar but a bit different. Here is the Wikipedia article on them (I'm surprised you guys have never come across this, perhaps it isn't used much outside of low-level Calculus): http://en.wikipedia.org/wiki/%28%CE%B5,_%CE%B4%29-definition_of_limit

Thanks for the reply Bob. I ended up getting a = 5. Is that the one and only solution to the problem?

And also, why did we want to make y a linear combination of the other three vectors?

I thought about subbing in 1 = h as it's no longer h+1 which gives 105 and it divides evenly by 7. However, that would just be like testing h = 2 from the very start, which is not how a proof should go (I think).

Edit: This website did it in the same way that I did and they have the solution: http://mathcentral.uregina.ca/QQ/database/QQ.09.99/pax1.html at the bottom.

However, I fail to understand the portion about the assumption.

That is probably it but that would mean he is suggesting precisely what we suggested: a number between 0 and 1. I guess we're all right.

Just an update, the TA said something practically along the lines of what you mentioned, Bobby.

Keep the A value on the side that it is on. Solve the inequality of A > 0. Then we solve the WHOLE inequality by isolating A. If B is between 0 and 1, then we simply switch the inequality sign, if not, we leave the sign as it is. Once that inequality is solved, we look at the inequality we solved earlier (A > 0) and conjure a solution by using methods such as a number line or whatever.

Those were her words, which seems quite on par with what you said I think.

Alright then. Thank you for all the help Bobby. Even if it may not be as clear as possible (as you put it), it still made sense and cleared up the problem for me.

I'll ask my TA sometime this week and post the answer.

.

That's a good way of understanding it. Would you say that the following generalizations are correct?
logb a > c  [assuming that the log part is on the left]
then we keep the a on the left side and get:

a > b^c

If b is 0<x<1:
then we switch the sign and get

a < b^c

Sorry Bobby, I was making a few edits there (taking out some of the stuff that I was able to get).

From what part of the solution did you get the 2^x < 1? The last part with the -2 < 2^x < 1? If so, I was able to understand how to get that part by splitting the inequality into two and solving.

However, the other question remains.