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! 
 A: Note that
$$\frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}=\frac{\sqrt{x^3}}{\sqrt{x^{3}}}\frac{1+\sqrt[6]{\frac{(x+1)^2}{x^9}}}{\sqrt{1-1/x^3}+1/\sqrt x}\to \frac{1+\sqrt{0}}{\sqrt{1-0}+0}$$
A: 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 $$\frac{x^{3/2}}{x^{3/2}}.$$
A: $$\lim_{x \rightarrow \infty}\frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$$
Using the "divide top and bottom by the highest power" method, the expression simplifies:
Top first term:
$$x\sqrt{x}=x^{3/2};\frac{x^{3/2}}{x^{3/2}}=1$$
Top second term:
$$\sqrt[3]{x+1}=(x+1)^{1/3}$$
$$\frac{(x+1)^{1/3}}{x^{3/2}}=\frac{(x+1)^{2/6}}{x^{9/6}}=\frac{(x^2+2x+1)^{1/6}}{x^{9/6}}=(\frac{x^2}{x^9}+\frac{2x}{x^9}+\frac{1}{x^9})^{1/6}=(\frac{1}{x^7}+\frac{2}{x^8}+\frac{1}{x^9})^{1/6}$$
Bottom first term:
$$\frac{(x^3-1)^{1/2}}{x^{3/2}}=(\frac{x^3-1}{x^3})^{1/2}=(1-\frac{1}{x^3})^{1/2}$$
Bottom second term:
$$\frac{x}{x^{3/2}}=\frac{x^{1/2}x^{1/2}}{x^{3/2}}=\frac{1}{x^{1/2}}$$
Now we have
$$\lim_{x \rightarrow \infty}1=1$$
$$\lim_{x\rightarrow \infty}(\frac{1}{x^7}+\frac{2}{x^8}+\frac{1}{x^9})^{1/6}=0$$
$$\lim_{x\rightarrow \infty}(1-\frac{1}{x^3})^{1/2}=1$$
$$\lim_{x\rightarrow \infty}\frac{1}{x^{1/2}}=0$$
So
$$\frac{1+0}{1+0}=1$$
