# Very lost with l'Hospital's rule

1. Nov 6, 2007

### coverticus

1. The problem statement, all variables and given/known data
Evaluate lim x$$\rightarrow$$ infinity of ($$\frac{x}{x+1}$$)$$^{}x$$, state explicitly the type of the indeterminate form.

2. Relevant equations

3. The attempt at a solution
I somewhat understand how to use l'Hospital's rule when the form is 0/0, but the inf/inf throws me off completely.
f'(x) = (x/(x+1))^x *ln(x/(x+1))
but from there or maybe the start I'm lost.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 6, 2007

### Midy1420

whether its 0/0 or inf/inf you proceed the same way...take the derivative of the top function and the derivative of the bottom function and evaluate the limit again

3. Nov 6, 2007

### Dick

First you need to change the expression to a 0/0 form. Hint: Take the log first, then try to rearrange that into a 0/0 form.

4. Nov 6, 2007

### coverticus

I should have posted this sooner, but isn't it in the form 1^inf ?

5. Nov 6, 2007

### Avodyne

Yes, that's why you need to take the log first. Find the limit of the logarithm, then exponentiate to get the limit of the original expression.

6. Nov 6, 2007

### coverticus

so you would get e^limx->inf of xln(x/(x+1)) and then what?

7. Nov 6, 2007

### Dick

Uh, rearrange xln(x/(x+1)) into a 0/0 form and apply l'Hopital to find the limit.

8. Nov 6, 2007

### coverticus

ok I got that much, and I got x/(1/ln(x/(x+1))), but I'm lost as to evaluate that lim with l'Hospitals, would it be (1) / (1/ln(x/(x+1)))', if so I haven't the slightest as to how to get that derivative.

9. Nov 6, 2007

### Dick

It's a lot easier to write it as ln(x/(x+1))/(1/x). Use the chain rule to do the ln part.

10. Nov 6, 2007

### coverticus

hey thanks for all your help guys, i got the answer (e^-1) finally and it was for a quiz so i appreciate it.