# Limits

• Feb 24th 2010, 04:35 AM
ukrobo
Limits
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}\
$
• Feb 24th 2010, 05:24 AM
tonio
Quote:

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
• Feb 24th 2010, 05:52 AM
drumist
Quote:

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.
• Feb 24th 2010, 06:05 AM
tonio
Quote:

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
• Feb 24th 2010, 11:59 AM
ukrobo