# Limit as x -> infinity of a sine/cosine graph.

1. Feb 24, 2010

### einsteinoid

1. The problem statement, all variables and given/known data

Lim [2 + 3x + sin(x)] / [x + 2cos(x)]
(x->infinity)

2. Relevant equations

3. The attempt at a solution

My roommate asked me to help him solve this homework question, at first glance I noted the derivative to be:

[3 + cos(x)] / [1 - 2sin(x)]

Now, the next step I'm assuming would be to plug infinity in for x. The answer in the book says that this should reduce the equation to 3. This would mean that by substituting infinity for x, the cos(x)/2sin(x) has to reduce to 1. But if x is approaching infinity, wouldn't a sin or cos wave just go back and forth between -1 and 1 forever? How, mathematically speaking, is it justifiable to say that they'd reduce to 1?

Thanks.

2. Feb 24, 2010

### tt2348

use the squeeze theorem and think more conceptually.
sin(x),cos(x) is on [-1,1]
so this means 2+3x+(-1)=1+3x < or =2+3x+sin(x) < or = to 2+3x+(1)=3x+3
also x+2(-1)=x-2 < or = x+2cos(x) < or = x+2(1)=x+2
then by pairing a bigger numerator to a smaller denominator, and a smaller numerator to a bigger denominator
you get
$$\frac{1+3x}{x+2} \leq \frac{2+3x+\sin x}{x+2\cos x} \leq \frac{3x+3}{x-2}$$

3. Feb 24, 2010

### einsteinoid

Ahhh, the ol' squeeze theorem. I never think to apply that. Thanks a lot, that's a big help!

4. Feb 24, 2010

### vela

Staff Emeritus
No, it doesn't. You're assuming the second limit exists, which it does not.
It's not. Your intuition is telling you the second limit doesn't exist, which is correct.

That's not a problem with L'Hopital's rule though. The rule says if the limit of the derivatives exists, then the original limit is equal to the limit of the derivatives. Since the limit of the derivatives doesn't exist, the rule doesn't help in this case.

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