Evaluating $\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $ $$\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $$ 
What method should I use to evaluate it. I can't use the  ${a^3}$-${b^3}$ formula because it is positive. I also tried to separate limits and tried  multiplying  with  $\frac {\sqrt[3]{(1-x^3)^2}}{\sqrt[3]{(1-x^3)^2}}$ , but still didn't get an answer. I got -$\infty$, and everytime I am getting $\infty -\infty$ .

For $n\gt2$ we have
$$\left(x-{1\over x^2}\right)^3=x^3-3+{3\over x^3}-{1\over x^6}\lt x^3-3+{3\over4}\lt x^3-1$$
Hence, since $x^3-1\lt x^3$ is clear, we have
$$0\lt x-\sqrt[3]{x^3-1}\lt x-\left(x-{1\over x^2}\right)={1\over x^2}\to0\quad\text{as }x\to\infty$$
So by the Squeeze Theorem,
$$\lim_{x\to\infty}\left(x+\sqrt[3]{1-x^3}\right)=0$$
