How to compare the limitation of $\frac{\log_2(x)}{ x^{(1/5)}}$

Question

I want to compare the time complexity of $\log_2(x)$ and $x^{(1/5)}$ but find hard to simplify them, I used Demos (a website) to draw the lines of them and find $log_2(x)$ is always higher than $x^{(1/5)}$, however, I do not know how to prove it.

Here is my work (not sure if it is correct)

I use $\lim\limits_{x\to\infty}\dfrac{\log_2(x)}{x^{(1/5)}}$, if the result is close to infinity so that the $log_2(x)$ is growing faster than $x^{(1/5)}$.

here to tell it, so I find derivatives of them: $\lim\limits_{x\to\infty}\dfrac{(\log_2(x))'}{(x^{(\frac{1}{5})})'}$ = $\lim\limits_{x\to\infty}\dfrac{\frac{1}{x \cdot \ln 2}}{\frac{1}{5}\cdot x^{(-\frac{4}{5})}}$

then I stuck here, how should I continue the provement?

Answer 1

$$L=\lim\limits_{x\to\infty}\dfrac{\frac{1}{x \cdot \ln 2}}{\frac{1}{5}\cdot x^{(-\frac{4}{5})}}$$

From this step, you can simplify it first,

$$L=\frac{5}{\ln2}\cdot \lim\limits_{x\to\infty}\dfrac{1}{x\cdot x^{(-\frac{4}{5})}}=\frac{5}{\ln2}\cdot \lim\limits_{x\to\infty}\dfrac{1}{x^{\frac{1}{5}}}=0$$

This means actually $x^{1/5}$ is much larger than $\log_2 x$, as $x\to\infty$.

Answer 2

After your comments, instead of applying the L'Hôpital's rule, you can go faster .

Substituting $\thinspace x\mapsto 2^{5t}$, then you get,

$$ \begin{align}\lim_{x\to\infty}\frac{\log_2(x)}{x^{\frac 15}}&=\lim_{t\to\infty}\frac {5t}{2^t}\\ &=5\lim_{t\to\infty}\frac {t}{2^t}\\ &=0\thinspace .\end{align} $$

Because, the polynomial $t$ grows asymptotically slower than $2^t$ as $t\to \infty\thinspace. $ Therefore ,

$$\lim_{t\to\infty}\frac {t}{2^t}=0$$

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In general, let $P_n(x)$ be a polynomial with degree $n$ and $a>1$ . Then :

$$\lim_{x\to\infty}\frac {P_n(x)}{a^x}=0\thinspace .$$