Is L'Hopital's Rule Applicable to Natural Log Limits?

[MrBailey]

Hi all.
I'm slightly confused with the following limit prob:
\lim_{x\rightarrow \infty} \frac{(ln (x))^n}{x}
which I know = 0. (n is a positive integer)
It looks like it's of indeterminate form, that is
\frac{\infty}{\infty}
Using L'Hopital's, it looks like you get another indeterminate form:
\lim_{x\rightarrow \infty} \frac{n(ln (x))^{n-1}}{x}
...and so on and so on.
Is it correct to assume that, applying L'Hopital's n times, you'll eventually get:
\lim_{x\rightarrow \infty} \frac{n\cdot (n-1)\cdot (n-2) \cdot ... \cdot 1}{x}
which is equal to zero?
Thanks,
Bailey

 

[MrBailey]

I edited the above...hopefully it looks clearer.

Isn't the derivative of the natural log 1/x
and you keep on doing it "n" times...so you always end up with an "x" in the denominator when applying L'Hopital's rule...until you get 1/x with all of the "n" terms in the numerator.

Please correct me if I'm wrong.
Bailey

 

[1800bigk]

yes that's right. chain rule. no matter what positive integer n, in the end you will have some fixed number in the numerator but the denominator goes to infinity so it wins out and the limit will be zero.

 

[MrBailey]

Thanks...I wanted to make sure I was on the right track.

Best wishes,
bailey