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

$$\lim_{x\to\infty}\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}} =\lim_{x\to\infty}\dfrac{\dfrac1{x^{4/3}}+\left(1+\dfrac1{x^4}\right)^{1/3}}{\dfrac1{x^{4/3}}+\left(9+\dfrac3{x^8}\right)^{1/6}}=\dfrac1{9^{1/6}}$$
which is $<1$
Hence the given limit should converge to $0$
Keep in mind $$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$$
