# Show that $\lim_{n \to \infty} \left(1+\frac{a_n}{n}\right)^n= e^a$ when $a_n \to a$

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?

• 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. – K.defaoite Nov 25 '20 at 0:50
• I suppose this is more like $\lim_{n \to \infty}f_n(a_n)$ where $f_n(x) = \left(1+\frac{x}{n}\right)^n$ – EssentialAnonymity Nov 25 '20 at 0:53
• I assume two facts only. – Danny Pak-Keung Chan Nov 25 '20 at 1:21

## 4 Answers

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.

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.

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

$$a-\frac{\delta}{2} so $$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 Hence, $$\left|\left(1+\frac{a_{n}}{n}\right)^{n}-e^{a}\right|<2\varepsilon$$. This shows that $$\left(1+\frac{a_{n}}{n}\right)^{n}\rightarrow e^{a}.$$

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 $$\lim_{n \to \infty} \left(1+\frac{0}{n}\right)^n = 1=e^0$$.

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