What's the algebraic trick to evaluate $\lim_{x\rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$? $$\lim_{x \rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$$
I got the first half:
$$\frac{x\sqrt{x}}{\sqrt{x^{3}-1}+x}=\frac{x\sqrt{x}}{\sqrt{x^{3}(1-\frac{1}{x^3})}+x}=\frac{1}{\sqrt{1-\frac{1}{x^3}}+\frac{1}{x^2}}$$
which evaluates to$\frac{1}{1+0}$.
For the second term $\frac{\sqrt[3]{x+1}}{{\sqrt{x^{3}-1}+x}}$ I can't get the manipulation right. Help is much apreciated! 

In the numerator, the terms are of order $x^{3/2}$ and $x^{1/3}$, so that the first dominates (the terms are added, there is no cancellation). In the denominator, $x^{3/2}+x^1$.
So the expression is virtually $$\frac{x^{3/2}}{x^{3/2}}.$$
