<s>[INST] How to evaluate $\lim\limits _{x\to \infty }\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}}\right)^{5x+1}$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}}\right)^{5x+1}$$
I tried like that:
$$\lim _{x\to \infty }\left(e^{\left(5x+1\right)ln\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+2}}\right)}\right)$$
Then:
$$ln\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+2}}\right)=\frac{\sqrt[3]{x^4+1}-\sqrt[6]{9x^8+2}}{x+\sqrt[6]{9x^8+2}}$$
$$=(5x+1)\frac{\sqrt[3]{x^4+1}-\sqrt[6]{9x^8+2}}{x+\sqrt[6]{9x^8+2}}$$
$$=\frac{5x^2\sqrt[3]x(1-\sqrt[3]{3})}{x+x\sqrt[3]x\sqrt[3]3}$$
Now I don't understand a thing:
the result of the latter fraction is $-\infty$ so $e^{-\infty}=\color{red}0$?
I'm sure I missed a few step, help me. Thanks
 [/INST] I'd set $x=1/t^3$ (with $x>0$, which is not restrictive) so the base becomes a fraction with
$$
\frac{1}{t^3}+\sqrt[3]{\frac{1}{t^{12}}+1}=
\frac{t+\sqrt[3]{1+t^{12}}}{t^4}
$$
at the numerator and
$$
\frac{1}{t^3}+\sqrt[6]{\frac{9}{t^{24}}+3}=
\frac{t+\sqrt[6]{9+3t^{24}}}{t^4}
$$
at the denominator. So, taking the logarithm, you have
$$
\lim_{t\to0^+}
\frac{5+t^3}{t^3}\log\frac{t+\sqrt[3]{1+t^{12}}}{t+\sqrt[6]{9+3t^{24}}}
=-\infty
$$
because the fraction under the logarithm has limit $1/\sqrt[3]{3}<1$.
Thus your limit is the same as $\lim_{u\to-\infty}e^u=0$.
</s>