<s>[INST] Calculate $\lim_{x\rightarrow\infty}\frac{x^3}{3^x}$. What is wrong with my calculations? $\frac{x^3}{3^x}=\frac{\exp(3\cdot\ln(x))}{\exp(x\cdot\ln(3))}=\exp(3\ln(x)-x\ln(3))$
$3\ln(x)-x\ln(3)=\frac{9(\ln(x))^2+x^2(\ln(3))^2}{3\ln(x)+x\ln(3)}$
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 
 [/INST] 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:
$$3\ln(x)-x\ln(3)=\frac{9(\ln(x))^2-x^2(\ln(3))^2}{3\ln(x)+x\ln(3)}$$
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