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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

1 Answer

1 vote

Answer:

n = 9

Explanation:

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


\lim_(x\to \infty)(\log x^k)/(x^n)=0

The derivative


(\log (x^k))'=(kx^(k-1))/(x^k)=(k)/(x)

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

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


\lim_(x\to \infty)(\log x^k)/(x)=\lim_(x\to \infty)(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


(f(x))/(x^n)=(x^8log(x^3)+x^6log(x^5))/(x^n)=(log(x^3))/(x^(n-8))+(log(x^5))/(x^(n-6))

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


\lim_(x \to \infty)(f(x))/(x^n)=0

and f is
O(x^n)

On the other hand, if
n \leq 8 then


(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)(f(x))/(x^n)=\infty

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

User Hyuk
by
7.9k points

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