<s>[INST] 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 $\infty -\infty$ .
 [/INST] 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
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