Evaluate $\lim_ {n \to\infty} n (n^ {\frac {1} {n}}-2^ {\frac {1} {n}}) ^a$ as $a$ varies in the reals How would you evaluate $\lim_ {n \to\infty} n (n^ {\frac {1} {n}}-2^ {\frac {1} {n}}) ^a$ with $n$ an integer and $a$ a real parameter? I tried to apply the most common criteria and to compare this with some other easier functions but without success.
Thank you
 A: If $a\le 0$
$$n\left( n^\frac 1n - 2^\frac 1n\right)^a = \frac{n}{\left(n^\frac 1n - 2^\frac 1n\right)^{|a|}} > \frac{n}{\left(n^\frac 1n \right)^{|a|}} = n^{1-\frac 1n |a|} \to \infty$$
If $a>0$
$$n^\frac 1n = e^{\frac 1n \ln n} = 1+ \frac 1n \ln n+O\left(\frac{\ln^2 n}{n^2}\right)$$
$$2^\frac 1n = e^{\frac 1n \ln 2} = 1+ \frac 1n \ln 2+O\left(\frac{1}{n^2}\right)= 1+ \frac 1n \ln 2+O\left(\frac{\ln^2 n}{n^2}\right)$$
Since $\frac 1n$ dominates $\frac{\ln^2 n}{n^2}$, we have
$$n^\frac 1n - 2^\frac 1n=\frac 1n \ln n-\frac 1n \ln 2+O\left(\frac{\ln^2 n}{n^2}\right) > \frac{\ln n}{2n}, \text{ when } n \text{ is large} \tag 1$$
And
$$n^\frac 1n - 2^\frac 1n=\frac 1n \ln n-\frac 1n \ln 2+O\left(\frac{\ln^2 n}{n^2}\right) < \frac 1n \ln n \text{, when } n \text{ is large }\tag 2$$
Now if $a>1$,
$$\text{when } n \text{ is large, } n\left( n^\frac 1n - 2^\frac 1n\right)^a < \frac{(\ln n)^a}{n^{a-1}} \to 0 \text{ via L'Hôpital}$$
If $0 < a \le 1$
$$\text{when } n \text{ is large, } n\left( n^\frac 1n - 2^\frac 1n\right)^a > \frac{n^{1-a} (\ln n)^a}{2^a} \to \infty $$
Conclusion: The limit is $\infty$ if $a\le 1$, $0$ if $a>1$.
A: Now I wasn't able to figure out a closed form but maybe it has something to do with the idea that I will present here. We will try different values for the parameter a. If a is equal to $0$ then obviously you just have the limit $\lim_{n\to\infty}n$, which diverges to $\infty$. If you let a be in the interval $(0;1]$ then it will diverge to $\infty$ once more. The same goes for the interval $(-\infty;0)$. The only place where this limit would converge is for a in the interval $(1;\infty)$, where the resulting limit will attain a value of $0$. Hopefully this helps, although someone more profficient in maths will surelly offer you a better advice.
