# How is the limit definition of e, actually equal to e? [duplicate]

$$\lim_{n\to\infty} (1+\dfrac{1}{n})^n = e.$$

To me this seems like $\dfrac{1}{n}$ goes to $0$, then you get $1^\infty$, which is equal to $1$. So why is it $e$?

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## marked as duplicate by Daniel Rust, anorton, Antonio Vargas, Alexander Gruber♦May 7 at 15:58

Because $1^\infty\neq1$ –  Jika May 7 at 15:40
then mathematicians have been wrong for hundreds of years lol. 1*1*1*1 = 1 no matter how many times you do it –  SN1 May 7 at 15:41
@SN1 : "You disproved mathematics". Go and publish your "work" and good luck. –  Jika May 7 at 15:48
Why don't you check for yourself? $(1+1/10)^{10}\approx2.594$, $(1+1/100)^{100}\approx2.705$, $(1+1/1000)^{1000}\approx2.717$, and so on. You can verify these facts with basic arithmetic if you don't believe me. –  Rahul May 7 at 15:51
The question title is a bit absurd. –  Awesome May 7 at 15:59

Well for a start, whenever $n \geq 1,$ we have $(1 + \frac{1}{n})^{n} \geq 1 + n \times \frac{1}{n} =2,$ so that disposes of the possibility that the limit as $n \to \infty$ could be $1.$ This isn't a proper answer, but could be expanded to give a rigorous justification that the limit is the same as the sum of the series $\sum_{k=1}^{\infty} \frac{1}{k!}.$ I just wanted to point out the absurdity of claiming that the limit could be $1$.

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Suppose you take logarithms and do $$\lim_{n\to \infty}n\ln (1+\frac 1n)$$

Now we know that the logarithm is greater than zero for all positive $n$, and that $$\infty \cdot \text{ anything greater than zero} = \infty$$ so the limit goes to infinity? Is there anything wrong with that argument?

And what do you do with $$\lim_{n\to \infty}n\cdot\frac 1n$$

Sometimes you have to consider the whole expression together, rather than taking it to pieces.

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$1^\infty$ is not $1$. The real numbers do not contain an infinity; you cannot use the rule that $\lim_{x \to \infty}f(x)^{g(x)} = (\lim_{x \to \infty}f(x))^{\lim_{x \to \infty}g(x)}$ unless $f(x)$ and $g(x)$ have well-defined, real number limits.

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If the real numbers dont have an infinity then the premise is wrong since we're finding the limit of something going to undefined (infinity). I have disproved mathematics –  SN1 May 7 at 15:44
@SN1 there is a very specific definition of what a limit as $n$ tends to $\infty$ means - this does not mean that $\infty$ has to be a real number. –  Daniel Rust May 7 at 15:45
@SN1: Taking limits to infinity is an abuse of notation; it means something distinct from the regular definition of the limit notation. (It seems that you're trying to get a reaction out of us, rather than actually learn, so this will be my last post on this question.) –  user2357112 May 7 at 15:46
so you admit it is wrong –  SN1 May 7 at 15:47

the $r$th term $T_r$ is $$\left(\frac1n\right)^r\frac{n(n-1)(n-2)\cdots (n-r+1)}{r!}=\frac1{r!}\prod_{s=0}^{r-1}\left(1-\frac sn\right)$$

Setting $\displaystyle n\to\infty,T_r=\frac1{r!}$

Now $\displaystyle e^x=\sum_{m=0}^\infty\frac{x^m}{m!},x=1\implies\cdots$

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