# L'Hospital's Rule -X to the power of X

1. Jan 4, 2013

### Astrum

1. The problem statement, all variables and given/known data
$Lim_{x\rightarrow0^{+}}X^{x}$

2. Relevant equations

3. The attempt at a solution

$ln(x^{x})=xln(x)=\frac{ln(x)}{x^{-1}}$

Take the derivative of the top and bottom -

$\frac{x^{-1}}{-x^{-2}}=\frac{-x^{2}}{x}$

So, we're left with: $lim_{x\rightarrow0^{+}}-x$

So, the answer would be 0, right? Where di I mess up? My book says the answer should be 1.

2. Jan 4, 2013

### Staff: Mentor

Since you used the ln to both sides, I think you forgot the last step to go back to the original equation and plug in the limit answer zero and hence e^0 = 1

3. Jan 4, 2013

### Zondrina

You forgot to do something. You took the ln, now to balance it out you have to raise something ;)

4. Jan 5, 2013

### Astrum

Ah, so, we take e to the power of - $lim_{x\rightarrow0^{+}}-x$

So you cancel out the ln?

5. Jan 5, 2013

### Staff: Mentor

No, you started with x^x and then converted it to e^xlnx and then you evaluated the limit of xlnx using lhospitals rule

so that limit was zero hence you have e^0 which is 1 hence x^x as x approaches 0 is 1.

6. Jan 5, 2013

### Astrum

Understood. Thanks.

7. Jan 5, 2013

### HallsofIvy

Another way of saying this is that you started with xx but then changed to y= ln(xx)= x ln(x). So what you found was that y[/b[ goes to 0. Because ln(x) is a continuous function that is the same as saying that ln(L)= 0 where "L" is the limit you are actually seeking. And, of course, ln(1)= 0 so L= 1.

8. Jan 5, 2013

### Astrum

Well, I thought of it like this: $x=e^{ln(x)}$

This is ok, right?

9. Jan 5, 2013

### Staff: Mentor

Yes, x = e ^ lnx andtherefore x^x = e ^xlnx