The Euler's constant $e$ appears whenever an extremely small change (increment or decrement) is applied an extremely large number of times e.g. an increment of 1% (1.01) applied a 100 times. It is easy to understand this concept of applying a small change a large number of times using compound interest.

The formula for compount interest is given by,

\begin{align}
A = P\left(1 + \frac{r}{100.n}\right)^n
\end{align}

where $P$ is the principal, $r$ is the yearly rate of interest in percentage, $n$ is the number of compounding periods and $A$ is the total amount at the end of 1 year. Let, $P=1$ and $r=100$.

If the interest is compounded annually, *i.e.* $n=1$, then

$$ A = (1+1)^1 = 2 $$

If the interest is compounded semiannually, *i.e.* $n=2$, then

$$ A = \left(1+\frac{1}{2}\right)^2 = 2.25 $$

If the interest is compounded quarterly, *i.e.* $n=4$, then

$$ A = \left(1+\frac{1}{4}\right)^4 \approx 2.441 $$

If the interest is compounded monthly, *i.e.* $n=12$, then

$$ A = \left(1+\frac{1}{12}\right)^{12} \approx 2.613 $$

If the interest is compounded daily, *i.e.* $n=365$, then

$$ A = \left(1+\frac{1}{365}\right)^{365} \approx 2.714 $$

If the interest is compounded hourly, *i.e.* $n=365\times24$, then

$$ A = \left(1+\frac{1}{365\times24}\right)^{365\times24} \approx 2.71812 $$

If the interest is compounded every second, *i.e.* $n=365\times24\times3600$, then

$$ A = \left(1+\frac{1}{365\times24\times3600}\right)^{365\times24\times3600} \approx 2.71828 $$

In the above situations, we observe that an extremely small
change, say $(1+\frac{1}{n})$ is being applied an extremely large
number of times *i.e.* $(1 + \frac{1}{n})^n$. In the limiting case,
if the interest is compounded continuously, the maximum amount $A$
that can be obtained at the end of one year is given by

\begin{align}
A &= \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n \\
&= \lim_{n\to\infty} \left(\binom n0 + \binom n1 \frac{1}{n} + \binom n2 \frac{1}{n^2} + \binom n3 \frac{1}{n^3} + \dots \right) \\
&= \lim_{n\to\infty} \left(1 + \frac{1}{1!} + \frac{n(n-1)}{2!}\frac{1}{n^2} + \frac{n(n-1)(n-2)}{3!}\frac{1}{n^3} + \dots \right) \\
&= 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \\
&= e \\
&\approx 2.718 281 828
\end{align}

This behaviour of applying an extremely small change an extremely large number of times occurs in many natural phenomena such as charging of a capacitor, decay of radioactive isotopes etc. In such processes, the small change is generally proportional to