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 $\infty -\infty$ .
 A: Hint: express:
$$x+\sqrt[3]{1-x^3}=x-\sqrt[3]{x^3-1}.$$
A: HINT: You were near to a solution, but the expression for $a^3+b^3$ should be used. 
$$
a+b=\frac{a^3+b^3}{a^2-ab+b^2},
$$
which leads to final limit 0.
A: I usually suggest to make the substitution $x=1/t$, so the limit becomes
$$
\lim_{t\to0^+}\frac{1+\sqrt[3]{t^3-1}}{t}=
\lim_{t\to0^+}\frac{1-\sqrt[3]{1-t^3}}{t}=
\lim_{t\to0^+}\frac{1-1+\frac{1}{3}t^3+o(t^3)}{t}
$$
A: 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
$$
\lim_{x\to\infty} (x+\sqrt[3]{1-x^3})=\lim_{x\to\infty}\frac{1}{(x^2-x\sqrt[3]{1-x^3}+(1-x^3)^{2/3})}
=\lim_{x\to\infty}\frac{1}{x^2(1-(x^{-3}-1)^{1/3}+(x^{-3}-1)^{2/3})}=0.
$$
A: Hint : make the following change of variable $u = \frac{1}{x^3}$
the limit becomes : $$\lim_{u \to 0} \frac{1}{u^{\frac13}}(1 - (1 - u)^{\frac13})$$
and use the fact that near $0$ you have : $(1-u)^{\frac13} \sim 1-\frac13u$
A: 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,
$$\lim_{x\to\infty}\left(x+\sqrt[3]{1-x^3}\right)=0$$
A: How about without calculus?  Does $x+\sqrt[3]{1-x^3}$ ever cross the x axis?
$$\begin{align}
x+\sqrt[3]{1-x^3}&=0 \\
x&=-\sqrt[3]{1-x^3} \\
x&=\sqrt[3]{x^3-1} \\
x^3&=x^3-1 \\
0&=-1
\end{align}$$
It doesn't.  It's also defined for all $x$.  And one point on the curve $f(x)=x+\sqrt[3]{1-x^3}$ is  $f(1)=1\ge0$.  Therefore $x+\sqrt[3]{1-x^3} \geq 0 \quad \forall x$.  The line $y=0$ serves as the greatest lower bound for $f(x)$
The goal now being to bound the function from the top, let there be some arbitrary value, $a>0$, that we wish to find all $x$ such that $f(x)$ is less than $a$.  Firstly we need to know when $f(x)=a$
$$\begin{align}
x+\sqrt[3]{1-x^3}&=a \implies 1-x^3=(a-x)^3=a^3-3a^2x+3ax^2-x^3 \implies \\
0&=3ax^2-3a^2x+a^3-1 \implies a \neq 0 \quad \land \\ 
x&=\frac{3a^2 \pm D}{6a} =\frac{a}{2} \pm \frac{D}{6a}\\
&\quad \text{discriminant} \ D=9a^4-4(3a)(a^3-1)=(3a)(4-a^3) \\
&\quad D=0 \implies [a=0] \lor \left[a=\sqrt[3]{4}=2^\frac{2}{3}\right] \\
&\quad \quad [a \neq 0] \implies  \\
&\quad \quad [a=\sqrt[3]{4}] \implies x=\frac{\sqrt[3]{4}}{2} \pm 0=2^\frac{-1}{3}\\
&\quad \quad f(2^\frac{-1}{3})=2 \cdot 2^\frac{-1}{3} =2^\frac{2}{3} \ \text{which is possibly} \max{f(x)} \\
&\quad D>0 \implies 0<a<2^\frac{2}{3} \\
&\quad D<0 \implies a \in \lbrace (-\infty, 0) \lor (2^\frac{2}{3}, \infty) \rbrace \\
\end{align}$$
3 points


*

*Seeing that the total range of intersection is the $y$ interval $(0, 2^\frac{2}{3}]$, and how the only place where the line $y=a$ within this range intersects with $f(x)$ only $1$ time is when $x=2^\frac{-1}{3}$ i.e. when $a=2^\frac{2}{3}$, the maximum point of $f(x)$ is $(2^\frac{-1}{3},2^\frac{2}{3})$.

*On the $x$ interval $(2^\frac{-1}{3}, \infty)$,  $f(x)$ can either stay the same value or decrease, since there will be no increasing.  Our point $(1,1)$ from earlier shows that $f(x)$ decreases on this interval

*with no least upper bound, yet a greatest lower bound of $0$, the end behavior of $f(x)$ tends to $0$


Probably easier with calculus
