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.

Have you tried dividing the original numerator and denominator each by $x^2 = \sqrt{x^4} = \sqrt[3]{x^6}\,$? This is a good example where algebraic manipulations are easier to use than is using L'Hopital.
$$
\lim_{x \to +\infty}\frac{\sqrt{x^4+1}}{\sqrt[\large 3]{x^6+1}}\cdot \frac {1/\sqrt{x^4}}{1/\sqrt[3]{x^6}} =\lim_{x\to+\infty} \frac{\sqrt{1+1/x^4}}{\sqrt[\large 3]{1+1/x^6}}=\frac{\sqrt 1}{\sqrt[\large 3]{1}}=1
$$
