Question

LIM ln(1/x)^x
x--->0

Answer 1

\[\lim_{x \rightarrow 0}\ln(\frac{ 1 }{ x })^{x}\]

Answer 2

i changed it to xln(1/x), and i let x=1/x such that i ended up having lnx/x , when i did the whole thing, i got infinity, wolfram has 0!

Answer 3

If you let \(x=\dfrac{1}{x}\), then \(\dfrac{1}{x}\to\infty\) since \(x\to0\).

Answer 4

I am not sure about his idea, but what I thought was that\[\int\limits_{ }^{ }\frac{1}{u}~dx~=\ln(u)\]and thus,\[\frac{d}{dx}~\ln(u)~=\frac{1}{u}\]

Answer 5

it should be 1/x in the integral.

Answer 6

and ln(x)

Answer 7

but I think that I am wrong .

Answer 8

which means if we differentiate ln(1/x) we get x?

Answer 9

x times 1/x^2

Answer 10

which is i/x

Answer 11

1/x I mean

Answer 12

\[\lim_{x\to0}\ln\left(\frac{1}{x}\right)^x=\lim_{x\to\infty}\ln x^{1/x}=\lim_{x\to\infty}\frac{\ln x}{x}\]
L'Hopital's rule from here:
\[\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{x\to\infty}\frac{\frac{1}{x}}{1}=0\]

Answer 13

so you have to change the limit to infinity....

Answer 14

That's if you use the substitution you mentioned.

Answer 15

oh thanx!, now it all makes sense, thank you so much. do you have another method?

Answer 16

You can still compute the limit as \(x\to0\):
\[\lim_{x\to0}\ln\left(\frac{1}{x}\right)^x=\lim_{x\to0}\ln x^{-x}=-\lim_{x\to0}\frac{\ln x}{\frac{1}{x}}\]
Then using L'Hopital's
\[-\lim_{x\to0}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\to0}\frac{\frac{1}{x}}{\frac{1}{x^2}}=\lim_{x\to0}\frac{1}{\frac{1}{x}}=\lim_{x\to0}x=0\]

Answer 17

thanx @SithsAndGiggles and @SolomonZelman . i am still getting used to the indeterminate form of limits