Determine the limiting behaviour of $\lim_{x \to \infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$ Determine the limiting behaviour of $\lim_{x \to \infty}{\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$
Used L'Hopitals to get $\;\dfrac{(x^6+1)^{\frac{2}{3}}}{x^2 \sqrt{x^4+1}}$ but not sure what more i can do after that.

Dividing numerator and denominator by $x^2$ gives
$$\lim_{x \to +\infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}=\lim_{x\to+\infty} \frac{\sqrt{1+x^{-4}}}{\sqrt[3]{1+x^{-6}}}=\frac{\sqrt{1}}{\sqrt[3]{1}}=1.$$
