Need to find the limit of a ratio of 2 functions

Question

I am very new to calculus, limits and more complex mathematics in general and I really need some help.

I need to compare the growth rate of 2 functions and because of this I need to find the limit of ratio of these functions $\lim\limits_{n \to \infty} \frac{g(n)}{f(n)}$

The functions are as follows:

$f(n) = \sqrt{n^3log_{2}n}$

$g(n) = 10n\sqrt{n}+100n$

How would I go about doing this? I've seen other people use $k=\sqrt{n}$ as a substitute when doing this.

Thank you in advance!

Answer 1

Observe that $n\sqrt n=\sqrt{n^3}$. $$\frac{10n\sqrt{n}+100n}{\sqrt{n^3\log_{2}n}}=\frac{10\sqrt{n^3}}{\sqrt{n^3\log_{2}n}}+\frac{100n}{n\sqrt{n\log_{2}n}}=\frac{10}{\sqrt{\log_{2}n}}+\frac{100}{\sqrt{n\log_{2}n}}$$ As you can see in the last expression we only have $n$ in the denominator. So as $n\to\infty$ the limit goes to zero.

Answer 2

$$\lim_{n \to \infty} \frac{g(n)}{f(n)} = \lim _{n \to \infty} \frac{10n\sqrt{n}+100n}{\sqrt{n^3\log_{2}n}} = \lim _{n \to \infty}\frac{10n^{3/2}+100n}{n^{3/2}(\log_{2}n)^{1/2}} = \lim _{n \to \infty}\frac{10+100n^{-1/2}}{(\log_{2}n)^{1/2}} $$

The numerator tends to $10$ while the denominator tends to $\infty$ so...

Answer 3

$$\frac{f(n)}{g(n)}=\frac{n^{3/2}\,\sqrt{\log_2(n)}}{10n^{3/2}+100n}=\frac{\sqrt{\log_2(n)}}{10+100n^{-1/2}}\to\infty.$$