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You may have seen the following equation:

\begin{align}

\lim_{n \rightarrow \infty} \left( 1 +\frac{1}{n} \right)^n = \mathrm{e}.

\end{align}

which expresses the fundamental constant e as a limit. Wondering how that can be proven?

Consider this graph:

\begin{align}

\int_1^{x} \, \frac{1}{t} \mathrm{d}t = \ln x

\end{align}

Looking at the graph, you can see that the area under the \(\frac{1}{t}\) curve on the interval \((1,1+\frac{1}{n})\) is bounded by

\begin{align}

\frac{1}{n}\left( \frac{1}{1+\frac{1}{n}} \right) \leq \int_1^{1+\frac{1}{n}} \, \frac{1}{t} \mathrm{d}t \leq \frac{1}{n} \left( 1 \right)

\end{align}

Using the definition of \(\ln x\) above, we have

\begin{align}

\frac{1}{1+n} \leq \ln \left( 1+\frac{1}{n} \right) \leq \frac{1}{n}

\end{align}

Raising \(\mathrm{e}\) to the power of each term,

\begin{align}

\mathrm{e}^{(\frac{1}{1+n})} \leq 1+\frac{1}{n} \leq \mathrm{e}^{(\frac{1}{n})}

\end{align}

Raising each term to power \(n\),

\begin{align}

\mathrm{e}^{(\frac{n}{1+n})} \leq \left( 1+\frac{1}{n} \right)^n \leq \mathrm{e}

\end{align}

As n approaches infinity, the left hand term approaches \(\mathrm{e}\), therefore

\begin{align}

\lim_{n \rightarrow \infty} \left( 1 +\frac{1}{n} \right)^n = \mathrm{e}.

\end{align}

QED.