# Let $f(n) = 5n^4 + 3n^3 − 5$. Show that $f(n)$ is $Θ(n^4)$. [closed]

How should this be proved? I know that $$f(n)$$ is $$Θ(n)$$ if and only if $$f(n)$$ is $$O(g(n))$$ and $$f(n)$$ is $$Ω(g(n))$$ and that If $$f(n)=\theta(g(n))$$ then $$\lim_{n \to \infty}\frac{f(n)}{g(n)}$$ will be a constant. Whats the better way to solve this? I have to directly find the constants k, C1, and C2

## closed as off-topic by kingW3, Alexander Gruber♦Apr 29 at 1:39

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• limit of $f(n)/n^4$ as $n$ to infinity is a constant and that's pretty easy to find. – kingW3 Apr 21 at 22:26
• I dont think I'm allowed to answer this in terms of limits. I need to use the theorems above in terms of discrete math. – ionics Apr 21 at 22:31
• Do you mean that you need to explicitly find constants $c_1, c_2$ and $n_0$ such that $c_1 g(n)\le f(n)\le c_2 g(n)$ for all integers $n > n_0$? – Minus One-Twelfth Apr 21 at 22:43
• Yes! Thank you. – ionics Apr 21 at 22:49
• You should include the definition you use of the theta function. – kingW3 Apr 21 at 23:09

## 1 Answer

Note that for $$n>1$$, $$8n^4>5n^4+3n^3>5n^4+3n^3-5>5n^4-2>3n^4$$