Question

Limit of function with L'Hopitals (Question below)

Answer 1

Hi guys, I'm completely stuck with one question:

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

I know I have to do this with L'Hopitals, but I can't get it in an indeterminate form where I can actually apply L'Hopitals. The trusty f(x)=e^ln(f(x)) trick doesn't work here with the existing ln function so I have no idea where to go with this. Any ideas? Thanks.

Answer 2

derive top and bottom

Answer 3

The derivative of ln(x) will be given by the chain rule.
Take the derivative natural log first, this will get you, 1/(1/x)
which is just as same as x

Then multiply this times the derivative of the inside,
d/dx (1/x) = d/dx (x)^(-1) = -x^(-2) = -1/x^2

So the entire derivative of the top is going to be,
-x/x^2 which simplifies to -1/x.

Answer 4

\[\Large \lim_{x \rightarrow \infty}\frac{\ln(\frac{1}{x})}{x}=\]
\[\Large \lim_{x \rightarrow \infty}\frac{\frac{d}{dx}[~\ln(\frac{1}{x})~]}{\frac{d}{dx}[~x~]}\]

Answer 5

Thanks for the reply. I can differentiate it okay, but the part I'm struggling with is getting it as an indeterminate form so I'm actually allowed to differentiate this. I know 0/0 and ∞/∞ are both indeterminate forms, but is 0/∞? Because that's what we have here.

Answer 6

\[\Large \lim_{x \rightarrow \infty}\frac{-\frac{x}{x^2}}{1}=\Large \lim_{x \rightarrow \infty}-\frac{x}{x^2}=\Large \lim_{x \rightarrow \infty}~~-\frac{1}{x}\]

Answer 7

wait it IS indeterminate form.

Answer 8

The ln is indeterminate,
because you have ln(1/x) and when x approaches infinity (roughly)
then, the ln is 0

So it is already undefined.....

Answer 9

Thanks again, but my lecture notes say I need my equation in the form: 0/0, ∞/∞, 0*∞, 0^0, ∞^0, 1^∞ or ∞-∞ before I can apply L'Hopitals. I understand that ln(1/x) tends to 0 as x tends to infinity, but that gives us this form so I'm not technically allowed to use L'Hopitals:

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

Answer 10

Just multiply numerator and denominator by -1.

Answer 11

Thanks for the reply, but which fraction are you referring to? The original limit? I'm probably being thick here but I can't see how that would help.

Answer 12

Yes I am referring to original limit. So then you would get indeterminate form you want, from there, you can apply L'Hôpital's rule.

Answer 13

Wouldn't I just get 0/-∞? I think I've worked it out though:\[\lim_{x \rightarrow \infty}\frac{\ln(\frac{1}{x})}{x} = \lim_{x \rightarrow \infty}\frac{-\ln(x)}{x} \approx \frac{-\infty}{\infty}\]Which satisfies the conditions for L'Hopitals so I'm good to go with the differentiation. Thank you both for your help!

Answer 14

That's what I was trying to say:

\[\lim_{x \rightarrow \infty}\frac{\ln\left(\dfrac{1}{x}\right)}{x} ~\Longrightarrow~ \lim_{x \rightarrow \infty}\frac{-\ln\left(\dfrac{1}{x}\right)}{-x} \lim_{x \rightarrow \infty}\frac{\ln(x)}{-x} \approx \dfrac{\infty}{-\infty}=\dfrac{-\infty}{\infty}\]

Answer 15

geerky number 1, lol, of course times -1 on top and bottom...!

Answer 16

but you are posted neg. infinity over positive infinity not as the final answer I hope(?)

Answer 17

Despite my negligence, I still managed to get the correct answer:)

Answer 18

Of course not. I suggest to do that, just so we can get indeterminate form we need, to apply L'Hopital's rule.

Answer 19

Oh, sorry @geerky42, I hadn't even considered manipulating the ln(1/x) so the -1/-1 was a bit confusing to me.

@what, no no, don't worry. I just had to satisfy that condition before I could differentiate and find the actual limit :)

Answer 20

But, in theory, or for a future reference, would it be an indeterminate form, if it had an undefined numerator and infinity on the denominator?
(Like I did initially)

Answer 21

As far as I know, no. From the definition here:

http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

We need 0/0 or ±∞/±∞.

Answer 22

You do need 0/0 or infinity/infinity to apply L'Hopital's; other (indeterminate? undefined?) (e.g. 0/infty, infty/0) forms cannot have L'Hopital's applied.

Answer 23

You got zero, correct?

Answer 24

Yeah I did, thanks.