Limits

**ukrobo** · February 24th 2010, 04:35 AM

For all $k>=0$ show that $n^k \in o(n^a)$
where:
$ a=log2(n) $

I get to here but i am not sure were to go next:
$ \lim_{n \to \infty} \frac{n^k}{n^a}\ $

**tonio** · February 24th 2010, 05:24 AM

Originally Posted by ukrobo

For all $k>=0$ show that $n^k \in o(n^a)$
where:
$ a=log2(n) $

I get to here but i am not sure were to go next:
$ \lim_{n \to \infty} \frac{n^k}{n^a}\ $

What does $n^k\in o(n^a)$ mean, anyway? I know $n^k=o(n^a)$ , but belongs...?

Tonio

**drumist** · February 24th 2010, 05:52 AM

Originally Posted by tonio

What does $n^k\in o(n^a)$ mean, anyway? I know $n^k=o(n^a)$ , but belongs...?

Tonio

http://en.wikipedia.org/wiki/Big_O_notation#Equals_sign

Basically it means the same either way.

**tonio** · February 24th 2010, 06:05 AM

Originally Posted by drumist

Big O notation - Wikipedia, the free encyclopedia

Basically it means the same either way.

Ok then, thanx. But then as $a=\log_2 n =\frac{1}{\log_n 2}$ , we get $n^a=n^\frac{1}{\log_n2}$ ,and thus $n^k=o\left(n^\frac{1}{\log_n2}\right)\Longleftrigh tarrow \frac{n^k}{n^\frac{1}{\log_n2}}=n^{k-\frac{1}{\log_n2}}\xrightarrow[n\to\infty]{}0\Longleftrightarrow k-\frac{1}{\log_n2}<0\Longleftrightarrow k<\frac{1}{\log_n2}$ .

Tonio

**ukrobo** · February 24th 2010, 11:59 AM

I need to evaluate the expression as either 0 or a real number. Your equation seems to imply that the expression = 0 to evaulate properties of k. I need to show that the expression = 0. cheers!

**tonio** · February 24th 2010, 07:05 PM

Originally Posted by ukrobo

I need to evaluate the expression as either 0 or a real number. Your equation seems to imply that the expression = 0 to evaulate properties of k. I need to show that the expression = 0. cheers!

I don't understand what you want: I thought you want to show that limit in your first post equals zero and I just wrote what the condition on k must be for that to be true.

Tonio

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