1. ## Limits

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

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

2. Originally Posted by ukrobo
For all $\displaystyle k>=0$ show that $\displaystyle n^k \in o(n^a)$
where:
$\displaystyle a=log2(n)$

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

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

Tonio

3. Originally Posted by tonio
What does $\displaystyle n^k\in o(n^a)$ mean, anyway? I know $\displaystyle n^k=o(n^a)$ , but belongs...?

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

Basically it means the same either way.

4. Originally Posted by drumist
Big O notation - Wikipedia, the free encyclopedia

Basically it means the same either way.

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

Tonio