Is the Limit of x^(1/(1-x)) as x Approaches 1 Equal to 0 or 1?

[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.

 

[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

 

[quasar987]

I got infinity as the answer .

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

And

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

 

[HallsofIvy]

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.

 

[SomeRandomGuy]

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

 

[ReyChiquito]

You don't have to use L'Hopital

look closely to your limit

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

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

Much more elegant don't you think?

 

[dextercioby]

You don't have to use L'Hopital

look closely to your limit

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

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

Much more elegant don't you think?

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... and of a very known limit.

Daniel.

 

[iSamer]

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... and of a very known limit.

Daniel.

nice and simple !

 

[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[/size]

 

[dextercioby]

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[/size]

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.

 

[daster]

Nice solution, dextercioby!