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I have seen a proof that shows $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n = e^x$$ by looking at the Taylor series expansion of $\ln(1+x)$ at $x=0$.

To prove a theorem, my textbook uses the fact $$\lim_{n \to \infty} \left(1+\frac{a_n}{n}\right)^n = e^a$$ when $a_n \to a$.

How can I prove this?

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  • $\begingroup$ Assuming sufficient smoothness requirements, the function of a limit is the same as the limit of the function. There's already plenty discussion about this topic on MSE. $\endgroup$
    – K.defaoite
    Nov 25 '20 at 0:50
  • $\begingroup$ I suppose this is more like $\lim_{n \to \infty}f_n(a_n)$ where $f_n(x) = \left(1+\frac{x}{n}\right)^n$ $\endgroup$ Nov 25 '20 at 0:53
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    $\begingroup$ I assume two facts only. $\endgroup$ Nov 25 '20 at 1:21
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You can try the following:

\begin{align*} \lim_{n\to \infty}\left( 1+\frac{a_n}{n} \right)^n &= \lim_{n\to \infty}\left( 1+\frac{a}{n} + \frac{a_n-a}{n} \right)^n \newline &= \lim_{n\to \infty}\left( 1+\frac{a}{n} \right)^n + \lim_{n\to \infty}\frac{a_n-a}{n}(\textit{something with finite limit}) \newline &= e^a + 0 \newline &= e^a \end{align*}

To formalize this second line you can use Binomial Expansion.

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We assume the facts

  1. $\lim_{n\rightarrow\infty}(1+\frac{x}{n})^{n}=e^{x}$ for any $x\in\mathbb{R}$.

  2. Exponential function $x\mapsto e^x$ is continuuous.


Suppose that $a_{n}\rightarrow a$. Let $\varepsilon>0$ be given. Since the exponential function $x \mapsto e^x$ is continuous at $a$, there exists $\delta>0$ such that $\left|e^{a}-e^{x}\right|<\varepsilon$ whenever $x\in(a-\delta,a+\delta)$. Since $a_{n}\rightarrow a$, there exists $N_{1}\in\mathbb{N}$ such that $\left|a_{n}-a\right|<\frac{\delta}{2}$ whenever $n\geq N_{1}$. Choose $N_{2}\in\mathbb{N}$ such that $1+\frac{a-\delta/2}{n}>0$ whenever $n\geq N_{2}$. (This is possible because $1+\frac{a-\delta/2}{n}\rightarrow 1$ as $n\rightarrow\infty$.)

Choose $N_{3}\in\mathbb{N}$ such that $\left|\left(1+\frac{a+\delta/2}{n}\right)^{n}-e^{a+\delta/2}\right|<\varepsilon$ whenever $n\geq N_{3}$. Choose $N_{4}\in\mathbb{N}$ such that $\left|\left(1+\frac{a-\delta/2}{n}\right)^{n}-e^{a-\delta/2}\right|<\varepsilon$ whenever $n\geq N_{4}$. Let $N=\max(N_{1},N_{2},N_{3},N_{4})$. Let $n\geq N$ be arbitrary, then we have

$$ a-\frac{\delta}{2}<a_{n}<a+\frac{\delta}{2}, $$ so $$ 0<1+\frac{a-\frac{\delta}{2}}{n}<1+\frac{a_{n}}{n}<1+\frac{a+\frac{\delta}{2}}{n}. $$ Raising to the $n$-th power, we further have $$ (e^{a}-\varepsilon)-\varepsilon<e^{a-\frac{\delta}{2}}-\varepsilon<\left(1+\frac{a-\frac{\delta}{2}}{n}\right)^{n}<\left(1+\frac{a_{n}}{n}\right)^{n}<\left(1+\frac{a+\frac{\delta}{2}}{n}\right)^{n}<e^{a+\delta/2}+\varepsilon<(e^{a}+\varepsilon)+\varepsilon. $$ Hence, $\left|\left(1+\frac{a_{n}}{n}\right)^{n}-e^{a}\right|<2\varepsilon$. This shows that $\left(1+\frac{a_{n}}{n}\right)^{n}\rightarrow e^{a}.$

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Write$\left(1+\cfrac{a_n}{n}\right)^n$ as $\left(\left(1+\cfrac{a_n}{n}\right)^{\cfrac{n}{a_n}}\right)^{\large{a_n}}$.

We know $\lim_{u\to \infty}\left(1+\dfrac1u\right)^u=e$. therefor $\lim_{n\to \infty}\left(1+\cfrac{a_n}{n}\right)^{\dfrac{n}{a_n}}=e$. and the original limit equal to $e^a$

Edit: As @DannyPak-KeungChan mentioned it is possible that value of $a$ be equal to $0$. in this case we should plug in this value in the original limit to get $\lim_{n \to \infty} \left(1+\frac{0}{n}\right)^n = 1=e^0$.

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  • $\begingroup$ The proof fails if $a_n=0$ for infinitely many $n$ (This is possible if $a=0$.) $\endgroup$ Nov 25 '20 at 1:36
  • $\begingroup$ That case need to be handled separately. $\endgroup$ Nov 25 '20 at 1:37
  • $\begingroup$ @DannyPak-KeungChan: thanks I edited my answer. it. is this ok now? $\endgroup$
    – Soheil
    Nov 25 '20 at 1:56
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Sketch of proof:

Note that $|f_n(a_n) - f(a)| \leqslant |f_n(a_n) - f(a_n)| + |f(a_n) - f(a)|.$

Then use that $f_n(x) = \left(1+\frac{x}{n}\right)^n \to f(x) =e^x $ uniformly on any compact set (proved here) and the exponential function is continuous.

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