# Good maths or bad maths?

Standard

Not sure, so maybe someone who knows can tell me.

Following on from the last blog, can we show $e = \lim_{n \to 0}(1 +n)^{\frac{1}{n}}$?

Assume a constant, $e$ exists such that $\frac{d}{dx}e^x=e^x$, could $\lim_{n \to 0}(1 + n)^{\frac{1}{n}}$ give us this $e$?

$\frac{d}{dx}e^x = \lim_{\delta \to 0} \frac{e^{x+\delta} - e^x}{\delta} =^{\lim_{\delta \to 0}}_{\lim_{n \to 0}}\frac{(1 +n)^{\frac{x+\delta}{n}} - (1 +n)^{\frac{x}{n}}}{\delta}$

$= (1+n)^{\frac{x}{n}}\frac{(1+n)^{\frac{\delta}{n}}-1}{\delta}$

Edit: Professor Rubin (see comments) tells me that, as I feared, what follows is not supportable:

Now, and this is the bit I have most doubts over, as $\delta \to 0$ and $n \to 0$ then $\frac{\delta}{n} \to 1$ and $1 + n - 1 \to 0$ and assuming $\frac {0}{0} = 1$ then we are left with $(1 + n)^{\frac{x}{n}}$.

Hence $\lim_{n \to 0} \frac{d}{dy} (1 + n)^{\frac{x}{n}} = (1 + n)^{\frac{x}{n}}$ and therefore $\lim_{n \to 0} (1+ n)^{\frac{1}{n}} = e$, the constant we are seeking.

But there are a lot of assumptions in there: anybody able to tell me how valid they are?

## 2 thoughts on “Good maths or bad maths?”

1. The bit about delta/n going to 1 is not supportable. Replace delta with n^2 (which approaches zero when n does) and your argument proves that n approaches 1 as n approaches 0 (immediately curing the Greek debt crisis, among other things).

Interchanging the order of limits is also tricky business. For any positive y, the limit of x/y as x approaches 0 from above is 0; so the limit of that limit as y approaches 0 from above is 0. Swap the order of the limits, treat the limit as y approaches 0 of x/y as infinity for positive x, and you get infinity for the outer limit.