Question

the limit as x goes to 0 of (x^(-1))/((ln(x))^2)

Answer 1

0

Answer 2

\[\frac{1}{x (ln{x})^2} \]This limit DNE I think :-)

Answer 3

Can't be 0...

Answer 4

It approaches -infinity from 0-, +infinity from 0+

Answer 5

L'hopitals doesn't apply, does it? I can't remember if you're allowed to use L'hopital's in the case of
\[\frac{\infty}{-\infty}\]

Answer 6

\[
\lim_{x\to0^-}\dfrac{1}{x(\ln(x))^2}=-\infty\\
\lim_{x\to0^+}\dfrac{1}{x(\ln(x))^2}=\infty
\]

Answer 7

Ah yes. What he said.

Answer 8

It does not, right? The limit is of the form 1/infinity.

Answer 9

Typo 1/0*

Answer 10

sorry my mistake, the anwer is infinity, just misread 0 for infinity

Answer 11

The answer isn't infinity either, the two-sided limit doesn't exist.

Answer 12

Wait... I'm not sure I agree with the left limit. Does it have a left limit? What happens when you try to evaluate that logarithm...

Answer 13

I did on the complex plane. Still, the limit DNE even if we are working with only R, we don't have the two sided limit, IMO.

Answer 14

haha this is math, bmp. Your opinion doesn't enter into it.

Answer 15

I agree though. I'm inclined to say DNE.

Answer 16

I was stating my opinion for the reason. That might not be the explanation for it, anyway, on R. But I don't think this exists.

Answer 17

Don't mind me. I'm just being overly snarky.

Answer 18

No problem, mate. You are right, I have to avoid using IMO, it's just an old vice :-)

Answer 19

Could either of you tell me how you got to that because I know L.H is needed but I don't know how to apply it to this

Answer 20

Well the only way I see of applying L'Hopital's is like this:
First of all, it only makes sense to consider the right hand limit. ln(x) does not exist for x<0.

Don't rewrite the problem, but keep it as:
\[\frac{\frac{1}{x}}{(\ln{x})^2}\]
The reason for this is that it means the limit as x approaches 0 will be of indeterminate form infinity/infinity, which allows me to use L'Hopitals rule.

So now I just take the derivative of the top and bottom, which gives:
\[\frac{\frac{-1}{x^2}}{2\ln(x)*\frac{1}{x}}\]

At this point, I'd simplify that and apply L'Hopital's rule again.

Answer 21

That simplified gives:
\[\frac{\frac{-x}{x^2}}{2\ln(x)}= \frac{\frac{-1}{x}}{2\ln(x)}\]

Notice that as x approaches 0, this is of indeterminate form -infinity/-infinity. Apply L'hopital's rule to get:

\[\frac{\frac{1}{x^2}}{\frac{2}{x}} = \frac{x}{2x^2} = \frac{1}{2x}\]

Which clearly approaches infinity.

Answer 22

And finally, gasp for air.

Answer 23

It seems that, doing only for R, the limit approaches infinity then. :-) Well done @SmoothMath

Answer 24

=D