Question

Fine the smallest positive integer n so that

f(x) = x^8 * log(x^3) + x^6 * log(x^5)

is a big-O of x^n

Answer 1

Answer:

n = 9

Step-by-step explanation:

Let's first prove that for any constants k > 0, $n\geq 1$

$\lim_{x\to \infty}\frac{\log x^k}{x^n}=0$

The derivative

$(\log (x^k))'=\frac{kx^{k-1}}{x^k}=\frac{k}{x}$

and the derivative $(x^n)' = nx^{n-1}$

Now, applying L'Hôpital's rule we find that

$\lim_{x\to \infty}\frac{\log x^k}{x}=\lim_{x\to \infty}\frac{k}{nx^n}=0$

Now, let f be the function

$f(x)=x^8log(x^3)+x^6log(x^5)$

It is easy to see that  f(x) is $O(x^n)$ only if $n\geq 9$

If $n\geq 9$  

$\frac{f(x)}{x^n}=\frac{x^8log(x^3)+x^6log(x^5)}{x^n}=\frac{log(x^3)}{x^{n-8}}+\frac{log(x^5)}{x^{n-6}}$

but both n-8 and n-6 are greater than one, so

$\lim_{x \to \infty}\frac{f(x)}{x^n}=0$

and f is $O(x^n)$

On the other hand, if $n \leq 8$ then  

$\frac{f(x)}{x^n}=x^{8-n}log(x^3)+x^{6-n}log(x^5)$

but 8-n is greater or equal than one, so

$\lim_{x \to \infty}\frac{f(x)}{x^n}=\infty$

and so f(x) in not $O(x^n)$