# Trigonometric limits

1. Dec 27, 2009

### reedy

Im trying to find

lim (2x+cos x)/x as x --> inf.

I've started out by multiplying with the conjugate.

[(2x+cos x)(2x-cos x)] / x(2x-cos x) =

=(4x2-cos2x)/(2x2-x cos x)

But now I'm lost - any hints?

2. Dec 27, 2009

### JSuarez

Don't do that. Just expand (2x + cos x)/x = 2 + (cos x/x), then take the limit.

3. Dec 27, 2009

### reedy

But what do I do about cos x/x as x--> inf? That isn't a limit I'm familiar with. My sources say it's 0, but is there any proof?

4. Dec 27, 2009

### JSuarez

Remember that cos x is a bounded function; then what is the limit of 1/x?

5. Dec 27, 2009

### reedy

im not sure of what a bounded function is, but you made me think:

lim cos x / x = lim cos x * lim 1/x

since lim 1/x is 0, lim cos x / x should be 0 as well. right? great stuff - thanks!

6. Dec 27, 2009

### JSuarez

You almost got it. Just don't write lim cos(x), when x goes to infinity, because it doesn't exist. Try this instead: a bounded function is one that satisfies |f(x)| < M, for some M and all x in f's domain; cos(x) is an example: |cos(x)| <= 1, for all x in R.

Then there is a theorem that states that the limit of the product of a bounded function times another that goes to 0, must be 0 as well, and this is exactly what you have here.