 # Determining lim_(n->oo}(n^((1)/(n))-1)^n with only elementary math 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 Roselyn Daniel 2022-07-21 Answered
Determining $\underset{n\to \mathrm{\infty }}{lim}{\left({n}^{\frac{1}{n}}-1\right)}^{n}$ with only elementary math
I am trying to find this limit:
$\underset{n\to \mathrm{\infty }}{lim}{\left({n}^{\frac{1}{n}}-1\right)}^{n},$
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
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it eri1ti0m
For $n\ge 5$ we have, using the binomial theorem
${\left(1+\frac{1}{2}\right)}^{n}\ge 1+\frac{n}{2}+\frac{n\left(n-1\right)}{8}\ge 1+\frac{n}{2}+\frac{n\left(5-1\right)}{8}=1+n>n$
Thus $1+\frac{1}{2}\ge {n}^{1/n}$, or
$0\le \left({n}^{1/n}-1{\right)}^{n}\le \frac{1}{{2}^{n}}$
(the first inequality being immediate.) Tking the limit as $n$ tend to $\mathrm{\infty }$, we get
$\underset{n\to \mathrm{\infty }}{lim}\left({n}^{1/n}-1{\right)}^{n}=0.$

We have step-by-step solutions for your answer! anudoneddbv
Perhaps the simplest way is to go step by step. First, show that ${n}^{1/n}\to 1$ as $n\to \mathrm{\infty }$. You can then conclude that $0<{n}^{1/n}-1<\frac{1}{2}$ for large enough values of $n$, meaning that, when $n\to \mathrm{\infty },$ the limit must be $0$ (sandwich principle).

We have step-by-step solutions for your answer!