Question

Real analysis/advanced calculus
How to:
Prove that a limit of a product is the product of the limits.

Answer 1

Some definitions and shortcuts that I will use:
L to denote the limit as x approaches x_0
E to denote epsilon (some arbitrarily small number)
d1 to denote delta 1
d2 to denote delta 2

Answer 2

Finally, I'll use d to denote delta.
Proof:
Let L[f(x)]=A, and L[g(x)]=B. We need to prove that L[f(x)*g(x)]=A*B,
ie show that there is some d>0, s. t. 0<|x-x_0|<d => |f(x)*g(x)-A*B|<E for all E>0;

and we are given that:

|f(x)-A| can be made arbitrarily small if 0<|x-x_0|<d1, and
|f(x)-B| can be made arbitrarily small if 0<|x-x_0|<d2.

Now, actually getting to the proof:
|f(x)*g(x)-A*B|=|f(x)*g(x)-A*g(x)-A*B+A*g(x)|=
=|g(x){f(x)-A}+A{g(x)-B}| ≤ |g(x){f(x)-A}|+|A{g(x)-B}| (by using the triangle inequality).

I'm just going to pull out g(x) and A from the absolute value:
|g(x){f(x)-A}|+|A{g(x)-B}|=|g(x)||f(x)-A|+|A||g(x)-B|

Now, since we already know that |f(x)-A| and |g(x)-B| can be bounded, we must show that |g(x)| is bounded...

Answer 3

..., or that:
|g(x)|=|(g(x)-B)+B| ≤ |g(x)-B|+|B| (again by using the triangle inequality)

|g(x)-B|+|B| < 1 + |B| (Since |g(x)-B| can be made arbitrarily small, we can make it less than 1), when |x-x_0|<d1

Continuing from |f(x)*g(x)-A*B| ≤ |g(x)||f(x)-A|+|A||g(x)-B|:
|f(x)*g(x)-A*B| ≤ |g(x)||f(x)-A|+|A||g(x)-B| < (1+|B|)*|f(x)-A|+|A|*|g(x)-B|<E/(1+|B|+|A|), and let the new epsilon, be E/(1+|B|+|A|), thus |f(x)*g(x)-A*B| can be made arbitrarily small, when epsilon = E/(1+|B|+|A|), QED.

Answer 4

Anyone interested in the proof for the quotient of the limits?

Answer 5

Triangle inequality:
|a+b|≤|a|+|b|

Answer 6

that was a small thing which i just mugged up but buddy this was a very nice proof..good job ^_^

Answer 7

So did anyone follow?

Answer 8

i maybe just did..hmmn

Answer 9

This is a freshman level problem...Why wouldn't people be able to follow?

Answer 10

There aren't that many college kids here, and to take real analysis, wouldn't you have to go through Cal1-3, which is not freshman anymore, unless you're given this in Cal3

Answer 11

I did epsilon delta proofs in my calculus 1 class in college...I've also taught it to my freshman.

Answer 12

Oh really, they didn't even teach it to us in Calc 3, we were just doing that trial and error bs