A question about the mathematical constant e.

The reason is that not all small numbers are the same. In other words, there are different rates of convergence. Here $1/n$ and $100/n$ both converge to 0, but at different rates. In fact $100/n$ is 100 times larger than $1/n$, although both are very small numbers when $n$ is large. When we just look at their limits we do not see this difference, they both converge to 0. Equivalently both $$(1+\frac{1}{n}),\quad (1+\frac{100}{n})$$ converge to 1 as $n\to \infty$, although $1+\frac{1}{n}$ converges to 1 faster, but we do not see this difference when we just look at the limit. However when we raise them to power $n$, which is a very large number, this difference becomes noticable. So the difference of the above two expressions, i.e. $\frac{99}{n}$, goes to 0 and is very small, but even this small difference can become important when we raise the expressions to the very large power $n$. Using the binomial theorem we can also see $$ (1+\frac{100}{n})^n = (1+\frac{1}{n} +\frac{99}{n})^n \\ \; \\=(1+\frac{1}{n})^n +n(1+\frac{1}{n})^{n-1}(\frac{99}{n})+ \dots +n(1+\frac{1}{n})(\frac{99}{n})^{n-1}+ (\frac{99}{n})^n$$ And the difference $n(1+\frac{1}{n})^{n-1}(\frac{99}{n})+ \dots +n(1+\frac{1}{n})(\frac{99}{n})^{n-1}+ (\frac{99}{n})^n$ is so large as $n\to \infty$ that changes the limit from $e$ to $e^{100}$.