Derive limit definition of e from differential equation definition of exponential function Given $f'=f$ and $f(0)=1$, show that $f(1)=\lim_{n\to\infty}(1+1/n)^n$.
Please don't prove it by proving that $f(1)=e$, separately proving that the limit equals e, and thus concluding that they're the same; that is, don't work from the ends to meet in the middle. I'm looking for a conceptual link between the stated starting and end points.
 A: Hint: start by splitting $[0,1]$ into $n$ parts of equal length. Observe that
$$
f\left(\frac{1}{n}\right) - f(0) \approx f’(0) \cdot \frac{1}{n} \Leftrightarrow f\left(\frac{1}{n}\right) \approx f(0) \cdot \left(1+\frac{1}{n}\right) \\
f\left(\frac{2}{n}\right) - f\left(\frac{1}{n}\right) \approx f’\left(\frac{1}{n}\right) \cdot \frac{1}{n} \Leftrightarrow f\left(\frac{2}{n}\right) \approx f\left(\frac{1}{n}\right) \cdot \left(1+\frac{1}{n}\right) \Rightarrow \\
\Rightarrow
f\left(\frac{2}{n}\right) \approx f\left(0\right) \cdot \left(1+\frac{1}{n}\right)^2
$$
Continuing, $f(1) \approx f(0) \cdot \left(1+\frac{1}{n}\right)^n = \left(1+\frac{1}{n}\right)^n$.
A: Define $h(x) := \lim_{n \to \infty} (1 + x/n)^n$.
We have:
\begin{align}
h'(x) &= \frac{d}{dx} \lim_{n \to \infty} (1 + x/n)^n \\
&= \lim_{n \to \infty} \frac{d}{dx} (1 + x/n)^n \\
&= \lim_{n \to \infty} (1 + x/n)^{n-1} \\
&= \lim_{n \to \infty} (1 + x/n)^n \\
 &= h(x)
\end{align}
This shows that $h$ satisfies the same initital value problem with $f$, so $h(x) = f(x)$, and in particular, $f(1) = \lim_{n \to \infty} (1 + 1/n)^n$.
