From Morris Kline's "Calculus: An Intuitive and Physical Approach":
At the end of the derivation (which I understand up to this point), he gets to $$\frac{\Delta y}{\Delta x} = \frac{1}{x_0} \cdot \log \bigg(1 + \frac{\Delta x}{x_0} \bigg)^{\frac{x_0}{\Delta x}}$$
and then says: "The point in making this change is that if we now let $\frac{\Delta x}{x_0}$ be $t$, we can consider the simpler expression $(1 + t)^{\frac{1}{t}}$ and see what happens as $t$ approaches $0$".
Of course, the point is that $\lim_{t \to 0} (1+t)^{\frac{1}{t}} = e$, which I understand.
However, What confuses me is that the choice of $\frac{\Delta x}{x_0}$ for $t$ seems completely arbitrary and doesn't work the other way around.
After all, since we're just choosing to replace an expression with a simpler one, I feel that we should be able to say let $\frac{x_0}{\Delta x} = t$ and then see what happens with $\lim_{t \to 0} (1+\frac{1}{t})^t$; but this limit is $1$, not $e$.
I know that in this latter expression, if we let $t$ approach $\infty$ instead of $0$, then we do get to $e$. And I can see that the choice of $\frac{\Delta x}{x_0}$ for $t$ has to do with the fact that we're ultimately seeking $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ with $\Delta x$ in the denominator.
But I don't understand why/how we choose $t$ to be $\frac{\Delta x}{x_0}$ and not the other way around. Why does this choice of $t$ allow us to take the same limit but the other choice of $t$ doesn't?
$\Delta x / x_0$ is a constant ($1/x_0$) multiplied by $\Delta x$, which doesn't change the answer you get when you consider a "small quantity" $\Delta x$ or $t = \Delta x / x_0$. However if you put $t = x_0 / \Delta x$ this is not proportional to $\Delta x$ (it is proportional to $1/ \Delta x$) so you have to treat the limit differently (i.e., letting $t \to \infty$) in order to get the right answer