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We define e to be a number which satisfies the following condition

$$\lim _{a \to 0} \frac{e^a-1}{a}=1. $$

How did we arrive to the following from above equation

$$e=\lim _{n \to \infty} \bigg(1+\frac{1}{n}\bigg)^n ? $$

so that we get the value of n

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  • $\begingroup$ sorry I dont know how to use mathjx. so can anyone edit it for me $\endgroup$ – Jayanth Kumar Jan 27 '15 at 7:09
  • $\begingroup$ Next time follow this meta.math.stackexchange.com/questions/5020/… $\endgroup$ – mattos Jan 27 '15 at 7:12
  • $\begingroup$ You can click "edit" and see how some of the mathjx works. $\endgroup$ – user99914 Jan 27 '15 at 7:14
  • $\begingroup$ We didn't arrive at the second definition of $e$ from the first.. They are two ways of describing the exponential function. The first comes from the gradient of the exponential function. The second comes from asking what happens if we have continually compounding interest. $\endgroup$ – mattos Jan 27 '15 at 7:33
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    $\begingroup$ @Mattos Then I guess the question really asked is: "How can we prove that these two numbers are the same?" $\endgroup$ – Arthur Jan 27 '15 at 7:50
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The approach you mention is difficult, but possible. I have presented it in my blog post. The main steps are as follows:

1) Define $a^{b}$ (without using any logs or $e$) rigorously for $a > 0$ and any real $b$.

2) Show that $\lim_{a \to 0}\dfrac{x^{a} - 1}{a} = f(x)$ exists for all $x > 0$ and hence defines a function of $x$. This function is denoted by $\log x$ and $e$ is then a number such that $\log e = 1$.

3) With $\log x$ defined above we have the following properties: $\log(xy) = \log x + \log y, \log 1 = 0, \log(1/x) = -\log x, \log(x^{y}) = y\log x$.

4) $\lim_{x \to 0}\dfrac{\log(1 + x)}{x} = 1$

5) Putting $x = 1/n$ in above limit and get $\lim_{n \to \infty}\log\left(1 + \dfrac{1}{n}\right)^{n} = 1$ and noting that $\log e = 1$ we get $\lim_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^{n} = e$.

Proof of 1), 2) and 4) is hard but not too hard.

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It's easier to start from the first equation and prove that it's the same as the second one.

From $$L := \lim _{a \to 0} \frac{e^a-1}{a}$$ set $e^a - 1 = t$, so that $a = \ln(1 + t)$: $$L = \lim_{t \to 0} \frac{t}{\ln(1 + t)} = \lim_{t \to 0} \left(\frac{\ln(1 + t)}{t}\right)^{-1} = \lim_{t \to 0} \left(\ln\left[\left(1 + t\right)^{1/t}\right]\right)^{-1} = \left(\ln\left[\lim_{t \to 0} \left(1 + t\right)^{1/t}\right]\right)^{-1}$$

The last step is legit because both $1/x$ and $\ln x$ are continuous functions. Now, recalling that $L = 1$ (from your first equation) we have that $$\begin{align} \ln\left[\lim_{t \to 0}\left(1 + t\right)^{1/t}\right] &= 1\implies\\ \lim_{t \to 0}\left(1 + t\right)^{1/t} &= e \end{align}$$

Or, equivalently, with one last change of variable $$e=\lim _{x \to +\infty} \left(1+\frac1x\right)^x.$$

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  • $\begingroup$ You wrote $(1+t)^1$ instead of $(1+t)^{1/t}$ at one point. $\endgroup$ – Akiva Weinberger May 15 '15 at 9:27
  • $\begingroup$ @columbus8myhw Nice catch, thanks! $\endgroup$ – rubik May 15 '15 at 16:43

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