Difficult limit involving $n$-th root and $n$-th exponentiation I've encountered this limit on a book without solution and after two days of thinking still I am not able to come up with an answer.

$$\lim_{n\to+\infty}\left(\alpha k^{\frac 1n}-1\right)^n,\: k\in\mathbb R_{>0},\: \alpha\in\mathbb N_{>0}.$$

A full complete and detailed soultion is welcomed. Also hints of course are accepted.
I've tried binomial Formula and to simplify the expression but all has been in vain. 

First of all, you should prove (if you do not already know) that if $k>0$ then $\lim_{n\to\infty}k^{1/n}=1$. The you have three cases (in the following I assume that $\alpha>0$ is not necessarily an integer):


*

*$0<\alpha<2$: then $|\alpha\,k^{1/n}-1|\to|\alpha-1|<1$ and $\lim_{n\to\infty}\bigl(\alpha\,k^{1/n}-1\bigr)^n=0$.

*$\alpha>2$: then $\alpha\,k^{1/n}-1\to\alpha-1>1$ and $\lim_{n\to\infty}\bigl(\alpha\,k^{1/n}-1\bigr)^n=\infty$.

*$\alpha=2$: we have a limit of the form $1^\infty$. The case $k=1$ is trival. I will do the case $k>1$ and leave for you the case $0<k<1$. Taking logarithms we see that it is enough to compute
$$
\lim_{n\to\infty}n\ln\bigl(2\,k^{1/n}-1\bigr).
$$
We have
$$\begin{align*}
n\ln\bigl(2\,k^{1/n}-1\bigr)&=\ln\bigl(1+2(k^{1/n}-1)\bigr)\\
&\sim 2\,n\bigl(k^{1/n}-1\bigr)\\
&=2\,n\bigl(e^{\ln k/n}-1\bigr)\\
&\sim 2\ln k
\end{align*}$$
as $n\to\infty$. Thus
$$
\lim_{n\to\infty}\bigl(2\,k^{1/n}-1\bigr)^n=k^2\quad\forall k>1.
$$

