Convergence of a basic sequence I know the following sequence converges to $e^8$, but I can't figure out why to save my life.
$$a_n=\left(1+\frac{8}{n}\right)^n$$
The sequence clearly is a manipulation of the basic limit derivation of $e$, but I couldn't figure out how to use it to get an answer. Instead, I've tried applying L'Hopital's and natural logs to reach the solution but can't seem to manage.
I'd really appreciate some help.
EDIT: Here was an attempt
$$\lim_{n\to\infty}e^{\ln(1+8/n)^n}$$
$$\lim_{n\to\infty}e^{n\cdot\ln(\frac{n+8}{n})}$$
$$\lim_{n\to\infty}e^n+\frac{e^{\ln(n+8)}}{e^{\ln n}}$$
$$\lim_{n\to\infty}e^n+\frac{n+8}{n}$$
But that's clearly wrong.
 A: Hint: $\;\;a_{8n}=\left((1+\frac{1}{n})^{n}\right)^8$
A: Hint:
Just show the limit of the log is $8$. For that, you can use equivalents: $\ln(1+x)\sim_0 x$.
A: $$\log(a_n) = n\cdot\log\frac{n+8}{n} = n\int_{n}^{n+8}\frac{dt}{t}=\int_{0}^{8}\frac{dt}{1+\frac{t}{n}}\underset{n\to +\infty}{\large\longrightarrow}\int_{0}^{8}\,dt = \color{red}{8}.$$
A: I believe the simplest approach is to use the well known limit $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = e\tag{1}$$ Note that the standard proof of the above result is that the limit on LHS exists and lies between $2$ and $3$ and this limit is denoted by $e$. Further the existence is proved by showing that the sequence $a_{n} = \left(1 + \dfrac{1}{n}\right)^{n}$ is increasing and bounded above.
It is important to note here that in $(1)$ the variable $n$ takes only positive integer values. Many people just assume that $(1)$ is also true if $n$ takes real values so that $$\lim_{x \to \infty}\left(1 + \frac{1}{x}\right)^{x} = e\tag{2}$$ The result $(2)$ is true but it lies on a different level (in terms of complexity) compared to $(1)$ because it involves development of theory of irrational exponents.
Hence we will not use the result $(2)$ (otherwise this exercise is trivial) and instead try to use only result $(1)$.
Note that if $k$ is any fixed positive integer and $a_{n}$ is any sequence then $$\lim_{n \to \infty}a_{n + k} = \lim_{n \to \infty}a_{n}\tag{3}$$ so that the sequence $b_{n} = a_{n + k}$ has exactly the same behavior as sequence $a_{n}$ when $n \to \infty$.
Therefore we have $$\lim_{n \to \infty}\left(1 + \frac{1}{n + k}\right)^{n + k} = e\tag{4}$$ We can now observe that
\begin{align}
L &= \lim_{n \to \infty}\left(1 + \frac{8}{n}\right)^{n}\notag\\
&= \lim_{n \to \infty}\left(\frac{n + 8}{n}\right)^{n}\notag\\
&= \lim_{n \to \infty}\left(\prod_{k = 1}^{8}\frac{n + k}{n + k - 1}\right)^{n}\notag\\
&= \lim_{n \to \infty}\prod_{k = 1}^{8}\left(\frac{n + k}{n + k - 1}\right)^{n}\notag\\
&= \prod_{k = 1}^{8}\lim_{n \to \infty}\left(1 + \frac{1}{n + k - 1}\right)^{n}\notag\\
&= \prod_{k = 1}^{8}\lim_{n \to \infty}\dfrac{\left(1 + \dfrac{1}{n + k - 1}\right)^{n + k - 1}}{\left(1 + \dfrac{1}{n + k - 1}\right)^{k - 1}}\notag\\
&= \prod_{k = 1}^{8}\frac{e}{1}\notag\\
&= e^{8}\notag
\end{align}
It is easy to extend the above proof to show that $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = e^{x}\tag{5}$$ for all rational $x$ and further it can be proved that the limit exists for irrational $x$ also.
