# 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

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 $\lim_{u\to-\infty}e^u=0$.

$$\lim_{x\to\infty}\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}} =\lim_{x\to\infty}\dfrac{\dfrac1{x^{4/3}}+\left(1+\dfrac1{x^4}\right)^{1/3}}{\dfrac1{x^{4/3}}+\left(9+\dfrac3{x^8}\right)^{1/6}}=\dfrac1{9^{1/6}}$$

which is $<1$

Hence the given limit should converge to $0$

Keep in mind $$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$$

• Surely $x+\sqrt[3]{x^{4}+1}=x^{4/3}\left(x^{-1/3}+\left(1+x^{-4}\right)^{1/3}\right)$? – Kevin Jan 28 '16 at 11:16