# 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.