# Find limits for x to infinity with L'Hopital

1. Jan 12, 2009

### Fairy111

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

For the following function decide whether f(x) tends to a limit as x tends to infinity. If the limit exists find it.

2. Relevant equations

f(x)=[xsinx]/[x^2 +1]

3. The attempt at a solution

I thought about using L'Hopitals rule, so i got:

[sinx + xcosx]/[2x]

So, sinx will keep between 0 and 1, and the xcosx and the 2x would both tend to infinity maybe?

Im not really sure though.

2. Jan 12, 2009

### Dick

Re: limits

l'Hopital's rule is going to get you into trouble here. x^2+1 goes to infinity, but x*sin(x) does not. It's oscillates. So l'Hopital doesn't apply. You'll have to approach the limit in a more basic way. How about dividing numerator and denominator by x?

3. Jan 12, 2009

### Staff: Mentor

Re: limits

A different approach is to use the fact that for x>=0, -x <= xsinx <= x, which means that (again for x >= 0)
$$\frac{-x}{x^2 + 1} \leq \frac{x sin(x)}{x^2 + 1} \leq \frac{x}{x^2 + 1}$$

The outermost expressions both have limits as x approaches infinity, and you can exploit this fact to say something about the expression in the middle. This idea is referred to in some texts as the "squeeze" theorem or "squeeze-play" theorem.