Question: > 
> Let $$\vec{v} = \frac{\vec{r}}{|\vec{r}|^3},$$ $\vec{r}=(x,y,z)$.
>  Evaluate
> 
> 
> $$ \iint\limits\_S \vec{v} \cdot \vec{n} \, dS$$
> 
> 
> with $\vec{n} $ pointing to the exterior of the surface $S$ and $S$
>  being a closed and smooth surface which contais the origin at the
>  interior.
> 
> 
> Answer: $ 4 \pi$.
> 
> 
> Sugestion: Consider a sphere centered at the origin contained at the
>  interior of $S$.
> 
> 
> 

**My attempt**: For spheres centered at the origin with radius $a > 0$, the surface integral will always be $4 \pi$. Can we bound $S$ between 2 of those spheres and use properties of the integral? I mean,

$$ 4 \pi \leq \iint\limits\_S \vec{v} \cdot \vec{n} \, dS \leq 4 \pi?$$

Answer: 
another way could be for you to use a substitution $x+1=t^3$. Then as $x \to 0$, $t \to 1$. Use this and simplify.
Use L'Hôpital's rule:
$\lim\limits\_{x\rightarrow0} \left[ \frac {(x+1)^{1/3}-1}{x} \right]=\lim\limits\_{x\rightarrow0}\left[\frac{\frac{d}{dx}[(x+1)^{1/3}-1]}{\frac{d}{dx}[x]} \right]=\lim\limits\_{x\rightarrow0}\left[\frac{\frac{1}{3}(x+1)^{-2/3}-0}{1}\right]=\lim\limits\_{x\rightarrow0}\left[\frac{1}{3}(x+1)^{-2/3}\right]=...$