<s>[INST] 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.
 [/INST] $$lim_{x \to \infty} \frac{\sqrt{x^4 + 1}}{\sqrt[3]{x^6+1}}$$
Mutliply top and bottom by $\frac{1}{x^2}$
$$lim_{x \to \infty} \frac{\sqrt{x^4 + 1}}{\sqrt[3]{x^6+1}} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} = lim_{x \to \infty} \frac{\sqrt{\frac{x^4}{x^4} + \frac{1}{x^4}}}{\sqrt[3]{\frac{x^6}{x^6}+\frac{1}{x^6}}}$$
$$= lim_{x \to \infty} \frac{\sqrt{1 + \frac{1}{x^4}}}{\sqrt[3]{1+\frac{1}{x^6}}} = \frac{\sqrt{1}}{\sqrt[3]{1}} = 1$$
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