# L'Hospital's Rule

1. Jul 10, 2010

### phillyolly

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

lim x^(1/x)
x->inf

3. The attempt at a solution

So, as I understand, there are several ways of solving it.
One of them, is not by using l'Hospital Rule because it is not an undeterminate function. OR IS IT?
If we rewrite the function as:
e^[lnx * (1/x)]

We get exponents: 1/infinity=0 and lnx=infinity, multiply both, it's 0.

Hence, e^0=1

2. Jul 10, 2010

### Bohrok

Careful, you have the right answer, but you can't use infinity like a number and say 0 × ∞ = 0. If instead you had limx→∞ x/ln(x), that would be x→∞ and 1/lnx → 0, so 0 × ∞, but that limit isn't 0, it's infinity.

$$\lim_{x\rightarrow \infty} x^{1/x} = \lim_{x\rightarrow \infty} e^{\ln x \cdot (1/x)} = \lim_{x\rightarrow \infty} e^\frac{\ln x}{x} = e^{\lim_{x\rightarrow \infty} \frac{\ln x}{x}}$$

So now you need to find lim×→∞ (lnx)/x, which you just did.

3. Jul 10, 2010

### Staff: Mentor

This is an indeterminate form, but L'Hopital's Rule is not applicable for this particular indeterminate form.
[$\infty * 0]$ is another indeterminate form, so you can't say with certainty that the limit of the exponent is 0.
Let y = ln(x)/x. Now take the natural log of both sides, and then take the limit of both sides. Presumably your text has an example that uses this approach.

4. Jul 11, 2010

### phillyolly

Well, your last one is essentially the same as the original when lnx/x. it looks like an indeterminate function.

Last edited: Jul 11, 2010
5. Jul 11, 2010

### Bohrok

6. Jul 11, 2010

### phillyolly

Thank you!

7. Jul 11, 2010

### Staff: Mentor

8. Jul 11, 2010

### phillyolly

Well, this shouldn't be so complicated cause this is just a beginner calculus.

9. Jul 11, 2010

### phillyolly

OK, so what I do in this problem is I take a derivative twice or three times... in order to solve this indeterminate function?

10. Jul 11, 2010

### Bohrok

As long as it's 0/0 or ∞/∞, you can keep using l'Hopital's rule until you get a different form that may lead to a limit or infinity.

11. Jul 11, 2010

### nonequilibrium

Or for those who like it more artisanal without l'Hôpital, define:

$$a_n = n^{1/n} - 1$$

Note that it suffices to prove that $$a_n \to 0$$.

From definition & Newton's binomium: $$n = (a_n + 1)^n = \sum \left( \frac{n}{i} \right) a_n^i \geq \frac{n(n-1)}{2} a_n^2^$$

From which directly follows that:

$$\frac{2}{n-1} \geq a_n^2 \geq 0$$ QED

EDIT: oh just noticed how the original poster stressed the use of l'Hôpital, my apologies, this renders my post useless, wasn't aware

Last edited: Jul 11, 2010
12. Jul 11, 2010

### phillyolly

That is a very important note to make. That is what I was not sure of. Thank you very much.

13. Jul 11, 2010

### phillyolly

Thanks a lot for your help!