Through working with convergence of series I've stumbled upon many that involved comparing $\log(n)$ with $n^{\alpha}$. Just recently a certain problem book claims the following:
There exists a natural number $n_0$ such that for every $n\ge n_0$ the inequality is true: $$\log(n)< n^{\alpha}\ \ \text{for all $\alpha> 0$}$$
I tried graphing couple of examples. It's obvious that this is true for $\alpha\ge \frac{1}{2}$ but for anything lower it takes a really long time for $n^{\alpha}$ to surpass $\log{n}$. Also lets consider that $$\lim_{n\to \infty}{n^{\frac{1}{n}}}=1$$
So for smaller and smaller $\alpha$ eventually $n^{\alpha}$ will be equal to $1$.However $\log(n)$ will keep increasing and go all the way to $\infty$. So it seems that after all this inequality is not true for all $\alpha > 0$. What is the boundary here? After which $0<\alpha<1 \ , n^{\alpha}$ can never surpass $\log{n}$ (assuming my provided statements are correct).
First of all, I want to note that you are interchanging limits at a point where it is not allowed.
Then, note Davids comment which reformulates the question to something that is correct whereas your statement isn't. To see why the result is true, simply take a look at the derivatives of the functions: $\log'(x)=1/x$, $(x^\alpha)'=\alpha x^{\alpha - 1} $ and appy l'Hopitals rule to the function $$ \frac{\log x}{x^{\alpha}} $$ to conclude that that function converges to 0 as $x$ approaches infinity.