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 $\infty -\infty$ .

Use the identity
$$
a^3+b^3=(a+b)(a^2-ab+b^2)
$$
with $a=x$ and $b=\sqrt[3]{1-x^3}$, to get that
$$
1=(x+\sqrt[3]{1-x^3})(x^2-x\sqrt[3]{1-x^3}+(1-x^3)^{2/3}).
$$ 
Thus
$$
\lim_{x\to\infty} (x+\sqrt[3]{1-x^3})=\lim_{x\to\infty}\frac{1}{(x^2-x\sqrt[3]{1-x^3}+(1-x^3)^{2/3})}
=\lim_{x\to\infty}\frac{1}{x^2(1-(x^{-3}-1)^{1/3}+(x^{-3}-1)^{2/3})}=0.
$$
