# Help w/ limit

1. Dec 13, 2004

### SomeRandomGuy

lim as x approaches 1 of x raised to the quantity 1/1-x. When I first did it, I got 1 as my answer, but I have a strong feeling the answer is 0 also. I don't know how to get it, however. 1 raised to any power is 1, that is where I got the answer 1 from.

2. Dec 13, 2004

### fourier jr

try l'hopital's rule. take ln of your function, then find the limit, then exponentiate to undo the ln

edit: the answer i got is 1/e

Last edited: Dec 13, 2004
3. Dec 13, 2004

### quasar987

I got infinity as the answer :grumpy:.

$$x^{\frac{1}{1-x}} = e^{\frac{lnx}{1-x}}$$

And

$$\lim_{x \rightarrow 1} \frac{lnx}{1-x} = \lim_{x \rightarrow 1} \frac{1}{x(1-x)^2} = \infty$$

4. Dec 13, 2004

### HallsofIvy

Staff Emeritus
How did you get
$$\lim_{x \rightarrow 1} \frac{lnx}{1-x} = \lim_{x \rightarrow 1} \frac{1}{x(1-x)^2} = \infty$$
?
Using L'Hopital, you would take the limit of $\frac{1/x}{-1}$ as x-> 1 which is -1. The limit of $x^{\frac{1}{1-x}}$ is e-1 as fourierjr said.

5. Dec 13, 2004

### SomeRandomGuy

That makes sense... I never even thought yo take the ln of x or use L. Hospital's rule. Thanks guys

6. Dec 13, 2004

### ReyChiquito

You dont have to use L'Hopital

$$\lim_{x \rightarrow 1} \frac{\log x}{1-x} = \lim_{x \rightarrow 1} \frac{\log x-\log 1}{1-x}$$

wich is the definition of the derivative of $-\log x$ evaluated in $1$ so its obvious than the limit is $-1$.

Much more elegant dont you think?

Last edited: Dec 13, 2004
7. Dec 14, 2004

### dextercioby

Not than my version
$$lim_{x\rightarrow 1} x^{\frac{1}{1-x}} =lim_{y\rightarrow 0} (1-y)^{\frac{1}{y}}=\frac{1}{e}$$
,where i made use of a simple substituion... :tongue2: and of a very known limit.

Daniel.

8. Dec 14, 2004

### iSamer

nice and simple !!

9. Dec 14, 2004

### ReyChiquito

hey Daniel, can you remind me of how

$$\lim_{y\rightarrow 0} (1-y)^{\frac{1}{y}}$$

is calculated???? too lazy to open my book :P

10. Dec 14, 2004

### dextercioby

Kay,i'll make an exception:
$$\lim_{y\rightarrow 0}(1-y)^{\frac{1}{y}}=\lim_{x\rightarrow +\infty}(1-\frac{1}{x})^{x} =\frac{1}{e}$$

Daniel.

11. Dec 14, 2004

### daster

Nice solution, dextercioby!