Limit of an infinite Series

[zzinfinity]

I have a problem that reads
Assuming that Lim(as n approaches infinity) of 1/n^c = 0 if c is any positive constant, show that

Lim (as n approaches infinity) of ln(n)/n^c =0.

I just sort of understand the denmoniator will become much bigger than the numerator as n approaches infinity, but I don't really know how to show it. Any suggestions would be appreciated.
Thanks!!!

 

[mmm4444bot]

Are you familiar with limit rules ?

Here's the rule (with simplified notation) for the limit of a product of two functions f and g:

lim [f * g] = lim [f] * lim [g]

The product in your given limit is:

ln(n) * 1/n^c

Apply the rule.

Function f is ln(n), and function g is 1/n^c.

They gave you the value of lim[g], yes ?

Try to make friends with the multiplicative property of zero.

(Yes, I realize that the statement in blue is ambiguous, but it works all ways.)

If you still need help, please ask specific questions.

Cheers ~ Mark

 

[mmm4444bot]

Somebody pointed out to me that the given exercise is a limit as n approaches infinity.

I misread the exercise as n approaches c.

My approach above is not valid because it leads to infinity times 0, and we cannot do arithmetic with infinity.

Please excuse my goof.

 

[lookagain]

Assuming that Lim(as n approaches infinity) of 1/n^c = 0 if c is any positive constant, show that

Lim (as n approaches infinity) of ln(n)/n^c =0.

\(\displaystyle \lim_{n \to \infty}\frac{ln(n)}{n^c}**\)


It is in the form of \(\displaystyle \frac{\infty}{\infty}\)


I will use L'Hopital's Rule:

\(\displaystyle * \ * = \lim_{n \to \infty}\frac {\frac{1}{n}}{cn^{c - 1}} = \lim_{n \to \infty}\frac{1}{c(n)n^{c - 1}} = \lim_{n \to \infty}\frac{1}{cn^c} = \lim_{n \to \infty}(\frac{1}{c} \ \cdot \ \frac{1}{n^c}) = (\lim_{n \to \infty}\frac{1}{c}) (\lim_{n \to \infty}\frac{1}{n^c}) = \frac{1}{c} \ \cdot \ 0 \ = \ 0\)


This method is based on using a certain technique that I don't know if the student has had.