How to solve the limit $\lim_{x \to \infty} \frac{e^{\log(x)^{c_1}}}{2^{c_2 x}}$ $$
\lim_{x \to \infty} \frac{e^{\log(x)^{c_1}}}{2^{c_2 x}}
$$
I tried to use L'hopital rule by differentiating top and bottom, but it ended up getting more and more complicated. Here, $c_1$ and $c_2$ are constants.

Suppose $a,b>0$
$$\log \left(\underset{x\to \infty }{\text{lim}}\frac{e^{\log ^a(x)}}{2^{b x}}\right)=\underset{x\to \infty }{\text{lim}}\log\frac{e^{\log ^a(x)}}{2^{b x}}=\underset{x\to \infty }{\text{lim}}\left(\log e^{\log ^a(x)}-\log2^{bx}\right)=$$
$$=\underset{x\to \infty }{\text{lim}}\left(\log^a x-bx \log 2\right)=\underset{x\to \infty }{\text{lim}}x\left(\frac{\log^a x}{x}-b\log 2\right)=-\infty$$
As the $\log$ of the limit is $-\infty$ the limit is $0$, provided that $a,b>0$.
