# L'Hopital's Rule

1. Nov 12, 2007

### physstudent1

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

Use L'Hopital's rule to evaluate the limit:

lim (x/(x+1))^x
x>infinite

2. Relevant equations

3. The attempt at a solution

I put it into a logarithm first to make it the limit of xln(x/(x+1)) then I took the derivative and got [(1/x)-(1/(x+1))]/[x^-2]
but its still indeterminate and after taking the 2nd derivative it was as well

Last edited: Nov 12, 2007
2. Nov 12, 2007

### cse63146

my guess would be you have to use some log properties like log(a/b) = log(a) - log(b)

3. Nov 12, 2007

### physstudent1

i did do that to get to the derivative but the derivatives didn't help

4. Nov 12, 2007

### cse63146

You forgot to use the power rule since there are 2 functions

This is what I got (could be wrong)

= x(logx - log (x+1)
= logx - log(x+1) + x((1/x)-(1/x-1)) <---simplify
= logx - log(x+1) - (1/x-1)

I believe you need to take the second derivative and you should get the limit to be 0 (it's been about a year since I did diffrential calc)

5. Nov 12, 2007

### Kummer

Why use L'Hopital? Just look at its reciprocal function. Trivial.

6. Nov 12, 2007

### cse63146

To get more experience using it?

7. Nov 13, 2007

### HallsofIvy

Staff Emeritus
Why the derivative? L'Hospital's rule usually involves taking two derivatives separately! Write the logarithm as
$$\frac{ln(x)- ln(x+1)}{x^{-1}}$$
and apply L'Hopital's rule to that.