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I am an undergrad math student, and I am just looking for intuition or even just some reason why

$$\lim_{n\to\infty}\Big(1+\frac{x}{n}\Big)^n=e^x$$

I know you can justify the answer with L'Hospital's Rule or power series, but those don't illuminate why this would be true in a purely logical, mathematical sense.

Edit: I understand this is sort of difficult to answer and is very open ended. I think that I can clear it up a bit. Simply plugging in $n=\infty$ into the limit would, hypothetically, give $1^\infty$, which of course is an indeterminate form. So then why is this indeterminate, and what does that have to do with $e$? Or is this really just the definition of $e$?

Sam Frank
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4 Answers4

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You can get this easily from the elementary definition of $e$, if that's what you want. Note:

$$\lim_{n\to\infty}\bigg(1 + \frac{x}{n}\bigg)^{n}\ \stackrel{t= \frac{n}{x}}{=}\ \lim_{t\to \infty}\bigg(\bigg(1 + \frac{1}{t}\bigg)^{t}\bigg)^{x}=\bigg(\lim_{t\to \infty}\bigg(1 + \frac{1}{t}\bigg)^{t}\bigg)^{x} = \boxed{e^{x}}$$

Joshua Wang
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If you ask a mathematician what the number $e$ is and what makes it so special, he will either say it's properties in analysis, such as $$\frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x,\quad \int e^x\, \mathrm{d}x = e^x+c $$ or it's beatiful combination with $i,\pi,1$ and $0$ $$e^{i\pi}+1=0,$$ the famous Euler's identity, which is usually proven by the also beatiful power series for $e$ that Euler discovered: $$\sum_{n=0}^{\infty}\frac{1}{n!}.$$

None of these formulas is the one you stated. But still, this formula has it's own history. It is the definition of $e$, how it all started, how Euler discovered further properties. Below are two interesting YouTube videos that tell you everything about the definition of $e$:

  1. More maths (starts at 1:18)
  1. More history (Numberfile)

Addendum: I remember quite well how I asked myself the exact same question ;).

vitamin d
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You just have the show the limit of the natural log of the expression is $x$.

Indeed, from a high school limit, $$\ln\biggl(1+\frac xn\biggr)^{\!n}=n\ln\Bigl(1+\frac xn\Bigr)=x\,\underbrace{\frac{\ln\Bigl(1+\frac xn\Bigr)}{\frac xn}}_{\substack{\downarrow \rlap{\:n\to\infty}\\1}} $$ tends to $x$ as $n$ tends to $\infty$.

Bernard
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Alternative intuition

$e$ is the real number such that $\frac{d}{dx} e^x = e^x.$

$\frac{d}{dx} (1 + [x/n])^n = n \times (1 + [x/n])^{(n-1)} \times [1/n] = (1 + [x/n])^{(n-1)}.$

By binomial expansion: $\lim_{n\to\infty} (1 + [x/n])^n ~=~ \lim_{n\to\infty} (1 + [x/(n+1)])^n~=~ \lim_{n\to\infty} (1 + [x/n])^{(n-1)}.$

user2661923
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