# Limits using l'Hopital

1. Oct 26, 2014

### ironman

Hello (:

Can someone help me calcultating this limit (as x → ∞)

This is the function:

If I just kept deriving sin(e^x) nothing really happened.

So I tried using L'Hopital but partly, i guess:

so:
(1 or -1) × 1 + 0 + 0 / 0 + ∞ + ∞

The answer must be 1 (according to wolfram alpha) but i don't see how that can be...

2. Oct 26, 2014

### SteamKing

Staff Emeritus
You probably should re-examine how to apply L'Hopital's Rule. What you are doing makes no sense.

3. Oct 26, 2014

### ironman

Ah yes I know: so because you get '∞/∞' in the original function you can use l'Hopital so you get

But then what am I to do with the e^x cos (e^x) because nothing useful will come if I keep deriving that part. (as far as I can see...)

4. Oct 26, 2014

### HallsofIvy

Instead of using "L'Hopital's rule" at all, you should think about which terms dominate the numerator and demonimator. The numerator and denominator each consist of three terms but one of those terms increases much faster than the other two.

5. Oct 26, 2014

### ironman

Ahh now it makes sense I think.
So in the nominator 3^x is the fastest growing term, and in de denominator that would be ln(x)^3 (?)
So if you divide by those terms, both the fastest growing terms will become 1 and the others will become 0.

6. Oct 26, 2014

### Staff: Mentor

Yes
No, it's not.

7. Oct 26, 2014

### Ray Vickson

Your notation is not sufficiently precise: by $ln(x)^3$ do you mean $\ln(x^3)$ or $(\ln x)^3$? If I read it using strict mathematical parsing rules it would come out at the latter.

8. Oct 26, 2014

### ironman

Than it must be 3^x...
but (ln(x))^3 with x to infinity equals: e to the power something equals infinity therefore (e^∞)^3 which grows faster than 3^∞ , right?

9. Oct 26, 2014

### ironman

I'm sorry. I mean:

10. Oct 26, 2014

### Staff: Mentor

No, (ln(x))3 doesn't grow faster than 3x. You have several errors in what you wrote above, one of which is the incorrect way you replaced (ln(x))3.

11. Oct 26, 2014

### ironman

Ok thanks Mark44 :) And thanks for the help guys!

12. Oct 26, 2014

### vela

Staff Emeritus
Because sine is bounded between -1 and 1, you could deal with the sine term by saying
$$\frac{-1 + x^2 + 3^x}{(\ln x)^3+2^x+3^x} \le \frac{\sin(e^x) + x^2 + 3^x}{(\ln x)^3+2^x+3^x} \le \frac{1 + x^2 + 3^x}{(\ln x)^3+2^x+3^x}$$ and then using the squeeze theorem, but this would be the roundabout way of solving the problem.

Think about the log function, perhaps look at its graph. Is $\ln x$ bigger than or smaller than $x$ when $x>1$?

13. Oct 26, 2014

### Cendrier

A trick for limits is to use a graphing calculator. Desmos is free and nice. Other than sin(e^x) everything is is going to infinity as the limit goes to infinity. Now you have to figure out what is growing the fastest so you can divide everything else by it. You can either graph everything individually to help you see it or take the derivative of every individual term to figure out the fastest grower.

Last edited: Oct 26, 2014