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The definition of the number $e$ that's used in my textbook is $e = \lim\limits_{x\to \infty} (1+\frac {1}{x})^x$ which relates to compound interest.

But when trying to calculate the derivative of $e^x$, I encountered another definition of $e$: $$e = \lim\limits_{x\to 0} (1+x)^\frac{1}{x}$$

Now, if it were $x$ approached $0$ from the positive direction, it would've made sense to me since $\frac {1}{x}$ would approach infinity and $1+x$ would approach $1$ with the same "speed"(sorry I don't know the term here) as the first definition. But it's approaching $0$ from both sides.

My question is, why are these two equal?

EDIT

The reason my question isn't a duplicate of the suggested one is that mine isn't really about the limit $\lim\limits_{x\to \infty} (1+\frac {1}{x})^x$. It's about its alternative form. I totally understand how the former isn't equal to $1$.

Thank you so much in advance!

P.S: I graphed the function $(1+x)^\frac {1}{x}$ and can "see" the limit exists as $x$ approaches $0$ but I would like a non-visual proof.

Harish Chandra Rajpoot
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3 Answers3

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The following general result holds for any real number $y$. $$\lim_{n \to \infty} \left(1+\frac{y}{n}\right)^n = e^y$$

To prove this, note that the limit of the logarithm is $$\lim_{n \to \infty} \log((1+\frac{y}{n})^n) = \lim_{n \to \infty} \frac{\log(1+y/n)}{1/n} = \lim_{n \to \infty}\frac{y/(n^2(1+y/n))}{1/n^2} = \lim_{n \to \infty} \frac{y}{1+y/n} = y,$$ by l'Hôpital's rule, so the original limit is $e^y$. Note that this assumes that you have defined $e$ somehow, and that you have defined the logarithm as the inverse of the function $x \mapsto e^x$.

If you have already shown that your definition of $e$ is equivalent to $\sum_{k = 0}^\infty \frac{1}{k!}$, then a similar argument using binomial expansions can work as well.

In particular, this result implies both $\lim_{x \to \infty}\left(1 + \frac{1}{x}\right)^x = e$ and $\lim_{x \to \infty}\left(1 - \frac{1}{x}\right)^x = e^{-1}$, which is what you need to conclude the two-sided limit $\lim_{x \to 0} (1+x)^{1/x} = e$.

angryavian
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You should look at the binomial expansion

$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ... $

If you now substitute $n$ by $1/x$ you get

$(1 + x)^{1/x} = 1 + \frac{1}{x}x + \frac{\frac{1}{x}(\frac{1}{x}-1)}{2!}x^2 + \frac{\frac{1}{x}(\frac{1}{x}-1)(\frac{1}{x}-2)}{3!}x^3 + ... $

$= 1 + 1 + \frac{1(1-x)}{2!} + \frac{1(1-x)(1-2x)}{3!} + ...$

and hence

$\lim_{x\rightarrow 0}(1 + x)^{1/x} = \lim_{x\rightarrow 0} 1 + 1 + \frac{1(1-x)}{2!} + \frac{1(1-x)(1-2x)}{3!} + ...$

$ = \sum_{k = 0}^{\infty} \frac{1}{k!} = e$

Carolin Zöbelein
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You want to prove $$\lim_{x\to -\infty}(1+\frac1x)^x = e\\\lim_{x\to-\infty}(1+\frac1x)^x=\lim_{x\to\infty}(1-\frac1x)^{-x}=\lim_{x\to\infty}\frac{1}{(1-\frac1x)^x}=\frac1{\frac1e} = e$$ To get $$\lim_{x\to\infty}(1-\frac1x)^x = \frac1e$$ You can prove that $$\lim_{x\to\infty}(1+\frac1x)^x(1-\frac1x)^x = \lim_{x\to\infty}(1-\frac1{x^2})^x=1$$

The last limit follows because $$(1-\frac1x)\leq(1-\frac1{x^2})^x\leq1$$ Lower bound follows from Bernouli

kingW3
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