# Another Euler number proof

OK, back to the issue of Euler’s number and the proof that it is $\lim _{x \to \infty}(1 + \frac{1}{x})^x$. This is a proof based on, though expanded from, what I picked up from a rather excellent website here.

Once again we start from the proposition that there exists a number $e$ such that $\frac{d}{dx}e^x = e^x$ and hence $\frac{d}{dx}log_e(x) = \frac{1}{x}$ (see here for why the second follows from the first).

Here we go… $y = \lim_{x \to \infty}(1 + \frac{1}{x})^x$ $log_e(y) = log_e[\lim_{x \to \infty}(1 + \frac{1}{x})^x]$ $= \lim_{x \to \infty}[x log_e(1 +\frac{1}{x})]$

At this point we still have $\infty \times 0$, an indeterminate number, so we need to look for a determinate form.

So we take this as: $= \lim_{x \to \infty}[\frac{log_e(1 + \frac{1}{x})}{\frac{1}{x}}]$

Now we have $\frac{0}{0}$, another indeterminate, but we can also apply L’Hospital’s rule.

This states that, $\lim_{x \to \pm\infty}[\frac{\frac{d}{dx}f(x)}{\frac{d}{dx}g(x)} = \frac{f(x)}{g(x)}]$ (a proof to follow sometime).

Here $f(x) = log_e(1+\frac{1}{x})$, $g(x) = \frac{1}{x}$ and, of course, $\lim_{x \to \infty}\frac{f(x)}{g(x)}= log_e[\lim_{x \to \infty}(1+\frac{1}{x})^x]$.

So, using the chain rule, $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$, for the numerator, $y = log_e(u)$, $u=1+\frac{1}{x}$: so $\frac{d}{dx}log_e(1 + \frac{1}{x}) = \frac{1}{1 + \frac{1}{x}}\times-\frac{1}{x^2}$

And for the denominator $\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}$, giving for the whole thing, $\lim_{x \to \infty}[\frac{1}{1+\frac{1}{x}}] =1$

Hence $log_e[\lim_{x \to \infty}(1 + \frac{1}{x})^x] = 1$, so $e=\lim_{x \to \infty}(1 +\frac{1}{x})^x$.

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