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.
 A: If you really want to use L'Hospital's Rule, take the $6$-th power of the expression. We get
$$\frac{(x^4+1)^3}{(x^6+1)^2}.\tag{$1$}$$
Those nasty roots are gone.
Now expand and use L'Hospital's Rule a dozen times.  Better yet, imagine expanding and using L'Hospital's Rule a dozen times. The top expands to $x^{12}+ \text{lower degree terms}$. The bottom expands to $x^{12}+ \text{lower degree terms}$.  It we use L'Hospital's Rule $12$ times, all the lower degree stuff will die, and we will be left with $1$, in the cumbersome form $\frac{12!}{12!}$. So the limit is $1$. Thus the limit of the original expression is $\sqrt[6]{1}$. 
But then again, once we see that the leading terms of the two polynomials are the same, the limit is obvious. 
A: 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.
$$
\lim_{x \to +\infty}\frac{\sqrt{x^4+1}}{\sqrt[\large 3]{x^6+1}}\cdot \frac {1/\sqrt{x^4}}{1/\sqrt[3]{x^6}} =\lim_{x\to+\infty} \frac{\sqrt{1+1/x^4}}{\sqrt[\large 3]{1+1/x^6}}=\frac{\sqrt 1}{\sqrt[\large 3]{1}}=1
$$
A: HINT
$$\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}} = \dfrac{x^2 \sqrt{1+1/x^4}}{x^2 \sqrt[3]{1+1/x^6}} = \dfrac{\sqrt{1+1/x^4}}{\sqrt[3]{1+1/x^6}}$$
A: $$ {\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}\sim {\frac{\sqrt{x^4}}{\sqrt[3]{x^6}}}=\frac{x^2}{x^2}=1 $$
A: Dividing numerator and denominator by $x^2$ gives
$$\lim_{x \to +\infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}=\lim_{x\to+\infty} \frac{\sqrt{1+x^{-4}}}{\sqrt[3]{1+x^{-6}}}=\frac{\sqrt{1}}{\sqrt[3]{1}}=1.$$
A: $$lim_{x \to \infty} \frac{\sqrt{x^4 + 1}}{\sqrt[3]{x^6+1}}$$
Mutliply top and bottom by $\frac{1}{x^2}$
$$lim_{x \to \infty} \frac{\sqrt{x^4 + 1}}{\sqrt[3]{x^6+1}} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} = lim_{x \to \infty} \frac{\sqrt{\frac{x^4}{x^4} + \frac{1}{x^4}}}{\sqrt[3]{\frac{x^6}{x^6}+\frac{1}{x^6}}}$$
$$= lim_{x \to \infty} \frac{\sqrt{1 + \frac{1}{x^4}}}{\sqrt[3]{1+\frac{1}{x^6}}} = \frac{\sqrt{1}}{\sqrt[3]{1}} = 1$$
