# 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.

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What is $k$? ${}{}{}$ –  martini Oct 2 '12 at 13:19

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