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I've been looking at some proofs of derivatives of functions and the teacher who made it used this in his proof:

$$ \lim_{u \to 0}((1+u)^{\frac{1}{u}})=e\ $$

I can see that this is true by substituting small values for u but why is this true and how do I prove it?

I would thank you if you could explain on more of a basic level, since I am quite a beginner here :)

Thanks in advance!

PS Sorry for my sloppy Enlgish

bp99
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    Well, what have you tried so far, and what are you stuck on? This limit is often taken to be the definition of $e$ (or in the equivalent form with $u\to\infty$, though it's still necessary to prove its convergence. – anomaly Sep 19 '16 at 20:13
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    And, this. And finally, this. – Caleb Stanford Sep 19 '16 at 20:17
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    This is often used as an alternative definition of $e$ – Alex Sep 19 '16 at 20:22
  • How are you defining $e$? One definition of $e$ is the number such that $$\int_1^e \frac{1}{x},dx=1$$while another is $$e=\sum_{n=0}^\infty \frac{1}{n!}$$an yet a third is $$e=\lim_{n\to \infty}\left(1+\frac1n\right)^n$$ – Mark Viola Sep 19 '16 at 20:49

2 Answers2

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Using the continuity of $\ln(x)$ and by setting the limit equal to $L\in\Bbb R$ we have: $$\begin{align}\lim_{u\to 0}(1+u)^{\frac{1}{u}}&=L\\\lim_{u\to 0}\frac{1}{u}\ln(1+u)&=\ln(L)\\\lim_{u\to 0}\frac{1}{1+u}&=\ln(L)\\1&=\ln(L)\\L&=e\end{align}$$ Note that L'Hopital's rule was used on line 3.

Dave
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  • This seems to be a circular argument. How are you defining the logarithm function? And what is the definition of $e$ you're using here? – Mark Viola Sep 19 '16 at 20:47
  • Define the natural logarithm as the integral of the standard hyperbola – Dave Sep 19 '16 at 21:59
  • First, one needs to show that the limit of interest exists. – Mark Viola Sep 19 '16 at 22:18
  • That's where I assumed the limit is equal to L where L is some arbitrary place holder, and then we show L is finite. – Dave Sep 19 '16 at 22:20
  • This isn't quite rigorous, but could easily have been made so. – Mark Viola Sep 19 '16 at 22:21
  • Define $\log(x)$ as $\log(x)=\int_1^x \frac{1}{t},dt$, $x>0$, and $e$ as the number such that $\log(e)=1$. It is easy to show that the logarithm is continuous for $x>0$ and satisfies $\log(x^n)=n\log(x)$. Then, we have $$\log((1+u)^{1/u}) =\frac1u \int_{1}^{1+u}\frac{1}{t},dt\tag 1$$The MVT shows that the limit of $(1)$ is $1$. – Mark Viola Sep 19 '16 at 22:25
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Hint:

Compute the limit of the logarithm of the expression.

Bernard
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