Determining $\lim_{n\to\infty}\left(n^{\tfrac{1}{n}}-1\right)^n$ with only elementary math I am trying to find this limit:
$$\lim_{n\to\infty}\left(n^{\tfrac{1}{n}}-1\right)^n,$$
I tried using exponential function, but I see no way at the moment.
I am not allowed to use any kind of differentiation or other topics of advanced math, only induction and school math are possible.
Thank you
 A: For $n\geq 5$ we have, using the binomial theorem
$$
\left(1+\frac{1}{2}\right)^n\geq 1+\frac{n}{2}+\frac{n(n-1)}{8}\geq
1+\frac{n}{2}+\frac{n(5-1)}{8}=1+n>n
$$
Thus $1+\frac{1}{2}\geq n^{1/n}$, or
$$
 0\leq (n^{1/n}-1)^n\leq \frac{1}{2^n}
$$
(the first inequality being immediate.) Tking the limit as $n$ tend to $\infty$, we get
$$\lim_{n\to\infty}(n^{1/n}-1)^n=0.$$
A: Perhaps the simplest way is to go step by step. First, show that $n^{1/n}\to 1$ as $n\to \infty$. You can then conclude that $0<n^{1/n}-1<\frac12$ for large enough values of $n$, meaning that, when $n\to\infty,$ the limit must be $0$ (sandwich principle).
A: Let n take the values $10^k$, for $k=1,2,3$, etc. $n^{^\tfrac1n}$ then becomes $10^{^\tfrac k{10^k}}$, whose exponent decreases powerfully towards $0$, making the entire quantity approach $1$, and the difference draw closer to $0$. Now, imagine that $0$ as $\dfrac1t$ , when $t\to\infty$. We have $\displaystyle\lim_{n\to\infty}\lim_{t\to\infty}\dfrac1{t^n}=\lim_{u\to\infty}\dfrac1u=0$.
