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: 
Let's put $1 + x = y$ so that $y \to 1$ as $x \to 0$. Then we have $$\begin{aligned}L &= \lim\_{x \to 0}\frac{(1 + x)^{1/3} - 1}{x}\\
&= \lim\_{y \to 1}\frac{y^{1/3} - 1}{y - 1}\\
&= \lim\_{z \to 1}\frac{z - 1}{z^{3} - 1}\text{ by putting }y = z^{3}, z\to 1\\
&= \lim\_{z \to 1}\frac{1}{z^{2} + z + 1} = \frac{1}{3}\end{aligned}$$

Note that we can prove the more general result in the same manner (i.e. using algebraic manipulation) that $$\lim\_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag {1}$$ if $a > 0$ and $n$ is rational. The result holds even if the number $n$ is irrational, but then the proof is not based on algebraical manipulation but rather on theory of logarithm and exponential functions.

Some answers have used the differentiation technique to get the answer. However to prove that $(x^{n})' = nx^{n - 1}$ we do need to use the limit result (1) I have mentioned above so that the reasoning based on differentiation to calculate this is kind of circular and should be avoided unless the question is kind of multiple choice type (where you have to get the answer and not show the working).
Use L'Hôpital's rule:
$\lim\limits\_{x\rightarrow0} \left[ \frac {(x+1)^{1/3}-1}{x} \right]=\lim\limits\_{x\rightarrow0}\left[\frac{\frac{d}{dx}[(x+1)^{1/3}-1]}{\frac{d}{dx}[x]} \right]=\lim\limits\_{x\rightarrow0}\left[\frac{\frac{1}{3}(x+1)^{-2/3}-0}{1}\right]=\lim\limits\_{x\rightarrow0}\left[\frac{1}{3}(x+1)^{-2/3}\right]=...$