Why does the $\alpha$ parameter grow to $\infty$ in Pareto-distributed random numbers when changing the threshold? I noted a strange fact. Let X be a set of Pareto distributed random number with $\alpha$ and $x_{\min}$ defined a priori.  Now, let $\alpha'$ be the estimated value of the shape and $x_{\min}'$ be a new threshold fixed a posteriori. 
If $x_{\min}'\rightarrow \max(x)$ then $\alpha'\rightarrow +\infty$. 
where $\max(x)$ is the max value in the sample.
I will try to explain me better.
Look at this code Comparing Pareto fitting methods now imagine to add after the line 
hh1 <- (matrix(rpareto(100,alpha,0.1),ncol=1)) 

(where 0.1 is $x_{\min}$)
this piece of code 
hh1 <- subset(hh1,hh1 > xmin) 

Here xmin${}=x_{\min}'$ namely the point where I start the fit or if you prefer a cut-off. 
Of course $x_{\min}'$ cannot be greater then $\max(x)$. Now imagine to put the code in a loop in order to see the behavior of $\alpha'$ while you increment the cut-off $x_{\min}'$.  
What I would expect is when $x_{\min}'\rightarrow \max(x)$ then $\alpha'\rightarrow\alpha_0$ because they are independent. But this not happens and $\alpha'$ (the estimated) tend to infinity. Why do we have such a strange behavior during the estimation?  Well I understand the behavior in the MLE method, but the others?
The same happens with alpha-stable distributions when we try make a measure of the shape parameter in the tails.
 A: If $x_{\min}^{'}\to\infty$ then we also have $x_i\to\infty$ and the type of convergence is "sure" convergence.  But note that just because two numbers both diverge does not mean that their limiting ratio is $1$.  This becomes clear once it is recognised that $x_{\min}$ is a scale parameter.  This is because we have:
$$\log\left(\frac{x_i}{x_{\min}}\right)\sim \operatorname{Expo}(\alpha)
$$
Expo(·) is the exponential distribution.  Now because the distribution is independent of $x_{\min}$ this is also the limiting distribution as $x_{\min}\to\infty$.
So the estimated value for $\alpha$ could be anything.  as the sample size increases, it will converge to the true value.
Update
In response to the revised question, the limit you are actually asking for is $x_{\min}^{'}\to x_{\max}$ not to infinity.  Now you ask why the estimate for $\alpha$ is infinite in this case.  Well the reason is that this limit corresponds to using a sample such that all of the values are equal $x_1=x_2=\dots=x_n=x_{\min}^{'}=x_{\max}$.  But in this case, the likelihood is exactly fitted by a dirac delta function.  The wikipedia page states that:
$$\lim_{\alpha\to\infty}f(x\mid x_{\min}^{'},\alpha)=\delta(x-x_{\min}^{'})$$
So the MLE procedure is not breaking down, but it is quite properly doing what it should: fitting the data as hard as it can within the class of distributions you give it.  You do get a warning though as the mle has infinite variance for all finite samples
