Graph this function on your TI84!

The standard method is to divide each term by the highest power term in the denominator. In this case, it's x. We don't have to bother with absolute values because x is approaching positive infinity in this case. $\lim_{x \rightarrow \infty} \dfrac{\dfrac{\sqrt{x^{2}-3}}{x}}{\dfrac{\sqrt[3]{x^{3}+1}}{x}}$ $=\lim_{x \rightarrow \infty} \dfrac{\dfrac{\sqrt{x^{2}-3}}{\sqrt{x^{2}}}}{\dfrac{\sqrt[3]{x^{3}+1}}{\sqrt[3]{x^{3}}}}$ $=\lim_{x \rightarrow \infty} \dfrac{\sqrt{\dfrac{x^{2}}{x^{2}}-\dfrac{3}{x^{2}}}}{\sqrt[3]{\dfrac{x^{3}}{x^{3}}+\dfrac{1}{x^{3}}}}$ Cancel, and apply the rule that $\displaystyle \lim_{x \rightarrow \infty} \dfrac{k}{x^{n}}=0$ , where k is a constant. You're left with $\displaystyle \lim_{x \rightarrow \infty} \dfrac{1}{1}$ which is simply $\boxed{1}$

You may transform the function into $\sqrt[6]\frac{(x^2-3)^3}{(x^3+1)^2} = \sqrt[6]\frac{x^6-9x^4+27x^2-27}{x^6+2x^3+1}$ Evaluating limit will give $\sqrt[6]1 = 1$