Calculate $\lim_{x\rightarrow\infty}\frac{x^3}{3^x}$. What is wrong with my calculations?

Question

$\frac{x^3}{3^x}=\frac{\exp(3\cdot\ln(x))}{\exp(x\cdot\ln(3))}=\exp(3\ln(x)-x\ln(3))$

$3\ln(x)-x\ln(3)=\frac{9(\ln(x))^2+x^2(\ln(3))^2}{3\ln(x)+x\ln(3)}$

I will use L'Hospital

$\frac{\frac{18(\ln(x))}{x}+2(\ln(3))^2x}{\frac{3}{x}+\ln(3)}$

And this goes to infinity, for $x\rightarrow \infty$

Therefore

$3\ln(x)-x\ln(3)$ goes to $\infty$ and $\exp(3\ln(x)-x\ln(3))$ must also go to infinity

Answer 1

This is incorrect: $$3\ln(x)-x\ln(3)=\frac{9(\ln(x))^2+x^2(\ln(3))^2}{3\ln(x)+x\ln(3)}$$

There should be a minus sign in the numerator: $$3\ln(x)-x\ln(3)=\frac{9(\ln(x))^2-x^2(\ln(3))^2}{3\ln(x)+x\ln(3)}$$