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.

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. 
