Show that $\lim_{n\to\infty} n^{2\alpha-1}-n^{2\alpha-2}\neq0$, if $\alpha>\frac{1}{2}$ I need to show that
$$
\lim_{n\to\infty} n^{2\alpha-1}-n^{2\alpha-2}\neq0,
$$ if $\alpha>\frac{1}{2}$. 
Maybe this is easier to read:
$$
\frac{1}{n}n^{2\alpha}-\frac{1}{n^2}n^{2\alpha}.
$$
I can sort of argue that $\lim_{n\to\infty}\frac{1}{n}n^{2\alpha}=\infty$, because $2\alpha>0$, so $n^{2\alpha}\geq n$ for each $n\in\mathbb N$. But I can't really squeeze a proof out of this.
So what I basically need is that $n^{\lambda}$, with $\lambda>0$, goes "faster to infinity", than $\frac{1}{n}$ and $\frac{1}{n^2}$ go to zero. 
Is there a theorem I could use? Or any other way to look at it?
 A: Your first equation
does not make sense,
because you are subtracting
two identical quantities
and claiming that
their difference
is nonzero.
You then write
$\frac{1}{n^2}n^{2\alpha}-\frac{1}{n}n^{2\alpha}
$.
This is just
$n^{2\alpha -2} - n^{2\alpha -1}$
which is the same as
$n^{2\alpha -1}(\frac1{n} -1)
$.
The term
$(\frac1{n} -1)
\to -1$
as $n \to \infty$,
so the limit depends on
$n^{2\alpha -1}$.
If $2\alpha -1 > 0$,
or
$\alpha > \frac12$,
$n^{2\alpha -1}
\to \infty$.
If $2\alpha -1 < 0$,
or
$\alpha < \frac12$,
$n^{2\alpha -1}
\to 0$.
If $2\alpha -1 = 0$,
or
$\alpha = \frac12$,
$n^{2\alpha -1}
= 1$.
A: Let $\alpha>0$. You have
$$
n^{2\alpha-1}-n^{2\alpha-2} = n^{2\alpha-1}\left(1-\frac{1}{n}\right)
$$
The first factor goes to $\infty$ with $n$, as $2\alpha-1>0$. The second (in parenthesis) converges to $1$. This should be enough for you to conclude directly.
If not, just observe that, for $n\geq 2$,
$$
n^{2\alpha-1}\left(1-\frac{1}{n}\right) \geq n^{2\alpha-1}\left(1-\frac{1}{2}\right) = \frac{n^{2\alpha-1}}{2} \xrightarrow[n\to\infty]{}\infty
$$
