limit of ${(1+s_n)}^n$ is one when limit of $n{s}_n$ is zero

The inequalities for differences of powers may be used to show that ${lim}_{n\to \mathrm{\infty }}{(1+s_n)}^n=1$ when ${lim}_{n\to \mathrm{\infty }}n{s}_n=0$. This fact plays an important role in the development of the theory of the exponential function as a limit of powers.

To derive this limit, we bound $1+s_n$ using the inequalities for differences of powers.

$$n{s}_n\le {(1+s_n)}^n-1\le \frac{n{s}_n}{1-(n-1)s_n}$$

Since ${lim}_{n\to \mathrm{\infty }}n{s}_n=0$, there must exist $N$ such that when $n>N$. Hence, when $n>N$,

so, as $n\to \mathrm{\infty }$, we have ${(1+s_n)}^n\to 1$.

Title	limit of ${(1+s_n)}^n$ is one when limit of $n{s}_n$ is zero
Canonical name	LimitOf1SnnIsOneWhenLimitOfNSnIsZero
Date of creation	2013-03-22 15:48:55
Last modified on	2013-03-22 15:48:55
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	7
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 26D99