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Hi all.

I'm slightly confused with the following limit prob:

[tex]\lim_{x\rightarrow \infty} \frac{(ln (x))^n}{x}[/tex]

which I know = 0. (n is a positive integer)

It looks like it's of indeterminate form, that is

[tex]\frac{\infty}{\infty}[/tex]

Using L'Hopital's, it looks like you get another indeterminate form:

[tex]\lim_{x\rightarrow \infty} \frac{n(ln (x))^{n-1}}{x}[/tex]

...and so on and so on.

Is it correct to assume that, applying L'Hopital's n times, you'll eventually get:

[tex]\lim_{x\rightarrow \infty} \frac{n\cdot (n-1)\cdot (n-2) \cdot ... \cdot 1}{x}[/tex]

which is equal to zero?

Thanks,

Bailey

I'm slightly confused with the following limit prob:

[tex]\lim_{x\rightarrow \infty} \frac{(ln (x))^n}{x}[/tex]

which I know = 0. (n is a positive integer)

It looks like it's of indeterminate form, that is

[tex]\frac{\infty}{\infty}[/tex]

Using L'Hopital's, it looks like you get another indeterminate form:

[tex]\lim_{x\rightarrow \infty} \frac{n(ln (x))^{n-1}}{x}[/tex]

...and so on and so on.

Is it correct to assume that, applying L'Hopital's n times, you'll eventually get:

[tex]\lim_{x\rightarrow \infty} \frac{n\cdot (n-1)\cdot (n-2) \cdot ... \cdot 1}{x}[/tex]

which is equal to zero?

Thanks,

Bailey

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