Proof that cos x diverges

1. Aug 5, 2011

bazingga123

1. The problem statement, all variables and given/known data
Recall that a function G(x) has the limit L as x tends to 1, written
lim as x -> infinity
G(x) = L,
if for any epsilon > 0, there exists M > 0 so that if x > M , then
|G(x) L| < epsilon.
This means that the limit of G(x) as x tends to 1 does not exist if for
any L and positive M, there exists epsilon > 0 so that for some x > M ,
|G(x) L| >= epsilon.
Using this deﬁnition, prove that

integral sinx dx from 2 pi to infinity diverges.

2. Relevant equations

3. The attempt at a solution

currently i just have no idea how to start this question up.
knowing that the integral is equals to -cos x + C, while cosine value is bouncing between -1 and 1, how do i start the proof using that definition

2. Aug 5, 2011

flyingpig

I think I just saw this question on Yahoo Answers lol

3. Aug 5, 2011

ArcanaNoir

Use the fact that certain multiples of x will give -1 and others will give 1 for x>M, thus you can't pick an arbitrary epsilon>0. That is, you can't get as close as you like to a limit L for all values x>M.

4. Aug 5, 2011

HallsofIvy

Staff Emeritus
Arcana Noir is referring to the fact that $\lim_{x\to a} f(x)= L$ if and only if $\lim_{n\to\infty}f(x_n)= L$ for every sequence $\{a_n\}$ such that $\lim_{n\to\infty} a_n= a$.

Look at $a_n= n\pi$ and $a_n= (2n+1)\pi/2$.

5. Aug 5, 2011

ArcanaNoir

Thank you for clarifying that, HallsofIvy, I only just learned this sort of thing this week, and only in terms of a sequence. I have the ideas but not the technical form yet.
The phrase "as close as you like to L" I picked up back in calc I, and have always preferred it conceptually to "an arbitrary epsilon >0".

Last edited: Aug 5, 2011
6. Aug 5, 2011

stringy

I think also we need to be a bit more careful with our definition of a limit and its negation.

To say that a function G does not have ANY limit L as x tends towards infinity means that (1) for all L (2) there exists an $\epsilon>0$ such that (3) for all $M>0$ (4) there exists an x in the domain so that

$$x>M$$ and $$|G(x)-L|\geq \epsilon.$$

Notice the sequence of quantifiers: (1) for all, (2) there exists, (3) for all, (4) there exists. This sequence is very important and if you change it the definition falls apart. Remember that the negation of $\forall$ is $\exists$ and the negation of $\exists$ is $\forall$.

7. Aug 5, 2011

Harrisonized

The integral is -cos x evaluated from 2π to ∞. In other words, it's:

-cos(2π) + lim cos(r) as r approaches ∞.
=-1+cos(∞)

But cos(∞) is equal to the limit of cos(1/x) as x approaches 0. Make a power series for cos(x) and replace all the terms with 1/x and you'll see clearly that it diverges. Then you can apply your limit definition on the power series and show that it's impossible.

8. Aug 5, 2011

Redbelly98

Staff Emeritus
Moderator's note:

Now that the OP has received several hints, please give bazingga123 a chance to work on the problem and reply before offering further help.