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.