Limit question calculus

Question

Can someone help me understand how this limit $\lim\limits_{n \to \infty}((n+1)/n)^n$ becomes $(1)^\infty$ (undefined)? I understand the limit can be found with logs but I'm still stuck on how this part is undefined.

limit in question:

\begin{equation*} \lim_{n\to\infty}\left(\frac{n+1}{n}\right)^{n} = 1^{\infty} \end{equation*}

Answer 1

$ 1^\infty =1$ Where $1$ is exactly $1$.

But the all the problem happens when $1$ is little bit greater or less than exact $1$ (In formal language $\to 1$) , then $1^\infty$ is indeterminate form which means by the nature of infinty you can get any limit for $1^{\infty}$.

for instance, $\lim_{n\to\infty}\left(\frac{n-1}{n}\right)^{n} = \frac{1}{e}$ ( Which is also in $1^\infty$ form)

You can verify the facts numerically by taking $n=10^{10}$

Answer 2

If I understand your confusion correctly, The confusion lies in $1^{\infty}$ part, When we say $1^n=1$ Which is of form $a^n$ for all Real values of "$n$" ,We absolutely know $a=1$ exactly

While talking About limits

When We say $(\rightarrow1)^{\rightarrow \infty}$

Now consider

$(\rightarrow1^{+})^{\rightarrow \infty}$

Let us see what binomial expansion tells us

$(1+\frac{1}{n})^n=\binom{n}{0}+\binom{n}{1}\frac{1}{n}+\binom{n}{2}(\frac{1}{n})^2.....$

Now $\binom{n}{0}$ and $\binom{n}{1}\frac{1}{n}$ are obvious determinate forms when $n\rightarrow\infty$,

All other sums like

$\binom{n}{2}\frac{1}{n^2}=\frac{n(n-1)}{2!(n^2)}$ and $\binom{n}{3}\frac{1}{n^3}=\frac{n(n-1)(n-2)}{3!(n^3)}$ etc. should be evaluated individually by applying limits

which means all $(\rightarrow1^{+})^{\rightarrow \infty}$ cases are not same, because each will have different binomial or other expansions

A quick demonstration of individual application of limit for above series is as follows

$L=\displaystyle \lim_{n \to \infty}\frac{n(n-1)}{2!(n^2)}=\displaystyle \lim_{n \to \infty}\frac{n^2(1-\frac{1}{n})}{2!(n^2)}=\frac{1}{2!}$

similarly doing such evaluations

you will get that

$\displaystyle \lim_{n \to \infty}(1+\frac{1}{n})^n=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}.........$

Now using expansion for $e^x$,Which can be proved without proving this limit, So that The argument is not circular and putting $x=1$ gives the same Summation obtained above

Hence the Limit $L=e$

Answer 3

((n+1)/n)=(1+1/n), when this is raised to the power of n it approaches e, the base of the natural logaritms, as n approaches infinity. This can be easily checked with a simple calculator, and it is very old knowledge.