What is the Order (Big O) of this polynomial?

Question

$$\frac{2n^{14} + 7 n^8 - 3}{3n^8 - n^4 + 3}$$

If this division is $p(n)$, I have to write $p(n) = O(n^k)$

I guess the answer is $n^6$, but how can i solve it step by step?

Answer 1

Step by step :

• $\frac {2n^{14} + 7 n^8 -3} {3n^8 - n^4 + 3}=\frac {n^{14}} {n^8}\frac {2 + 7 n^{-6} -3n^{-14}} {3 - n^{-4} + 3n^{-8}}$

• When $n\longrightarrow \infty$, the right side fraction has finite limit $\frac 2 3$, so it is bounded. Let $B\in\mathbb R$ be this bound.

• $\frac {2n^{14} + 7 n^8 -3} {3n^8 - n^4 + 3}=n^{6}\frac {2 + 7 n^{-6} -3n^{-14}} {3 - n^{-4} + 3n^{-8}}\leqslant n^6B$

Hence $O(n^6)$

Answer 2

Since $$\frac{2n^{14}+7n^8-3}{3n^8-n^4+3}=\frac{2n^6+7-\frac{3}{n^8}}{3-\frac{1}{n^4}+\frac{3}{n^8}}$$ it will be a $O(n^6)$.