# Extention of l'hopital's rule?

1. Aug 6, 2008

### avr10

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

Hi all, I'm new here, and just have a quick question:

Evalute: lim as x$$\rightarrow$$$$\infty$$ of $$\frac{x - sin x}{x^{3}}$$

2. Relevant equations

l'hopital's rule?

3. The attempt at a solution

So far I've used L'hopital's rule to get it down to lim as x$$\rightarrow$$$$\infty$$ of $$\frac{cos x}{6}$$

but after that, I don't know what to do... I'm pretty sure that I have to use an extention of L'hopital's rule

thanks!

2. Aug 6, 2008

### evilpostingmong

At infinite, cos(x)=1

3. Aug 6, 2008

### Dick

Don't be silly. cos(x) oscillates, it doesn't converge. And, arv10, if you are going to apply l'Hopital's rule multiple times, you have to make sure that at each stage you still have an infinity/infinity or 0/0 form. After the first application you have (1-cos(x))/(3x^2). That's not a form you can apply l'Hopital's rule to. You have to apply another technique to that. And once you've done that you should realize you didn't need to apply l'Hopital to begin with.

4. Aug 6, 2008

### buffordboy23

No. The squeeze theorem is your friend here.

5. Aug 7, 2008

### HallsofIvy

Staff Emeritus
Thought I had replied to this before.

You can't use L'Hopital's rule: The numerator does not go to either 0 or infinity: sin(x) oscilates between -1 and 1 and x goes to infinity.

But
$$\frac{x- sin x}{x^3}= \frac{x}{x^3}- \frac{sin x}{x^3}= \frac{1}{x^2}- \frac{sin x}{x^3}$$
The first fraction obviously goes to 0 and the second fraction, because sin x is always between -1 and 1, also goes to 0.

6. Aug 7, 2008

### Tobias Funke

While I agree that L'Hospital's rule isn't necessary in this problem, it's not required that the fraction be of the form $$\infty/\infty$$. All that's required is that the denominator approaches infinity, so you can try L'Hospital as far as avr10 did, although it tells you nothing at the end since what you end up with has no limit.

7. Aug 8, 2008

### HallsofIvy

Staff Emeritus
Which, since this does have a limit, means that L'Hopital's rule does NOT apply here.

8. Aug 8, 2008

### Tobias Funke

Well, it doesn't help you if you go all the way with it, but if for some reason you think the limit that Dick gets after using L'Hopital once is easier than the original, I would say it applies. Anyway, that's just semantics. I was just pointing out that the numerator not going to infinity doesn't rule out the limit from being a "candidate" to try L'Hopital.

9. Aug 8, 2008

### carlodelmundo

Correct me if I'm wrong (curious). But shouldn't the numerator "x - sin(x)" approach infinity? That is even if sin(x) oscilates from -1 to 1 ... the variable "x" is subtracted from "sin(x)" and as x increases.... the subtraction from (-1 or 1) is so miniscule that x approaches infinity?

Graphically--y = x - sin(x) resembles a cubic function. Please shed some light.

Carlo

P.S. - I agree that the limit is 0.

10. Aug 8, 2008

### Dick

You CAN apply l'Hopital to (x-sin(x))/x^3. You get (1-cos(x))/(3x^2). You CAN'T apply l'Hopital to that.

11. Aug 8, 2008

### Tobias Funke

Sure you can. The denominator goes to infinity. You don't need the form $$\infty/\infty$$ to apply the rule (check baby Rudin). However, that's the last time that application of the rule yields something with a limit, so if you have an easier time showing that sinx/(6x) approaches 0, then you've successfully used L'Hospital to solve the problem, albeit not in the shortest way.