# Exponential Limit of (1+1/n)^n=e

In this tutorial we shall discuss the very important formula of limits,  $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = e$

Let us consider the relation
${\left( {1 + \frac{1}{x}} \right)^x}$

We shall prove this formula with the help of binomial series expansion. We have
$\begin{gathered} {\left( {1 + \frac{1}{x}} \right)^x} = 1 + x\left( {\frac{1}{x}} \right) + \frac{{x\left( {x – 1} \right)}}{{2!}}{\left( {\frac{1}{x}} \right)^2} + \frac{{x\left( {x – 1} \right)\left( {x – 2} \right)}}{{3!}}{\left( {\frac{1}{x}} \right)^3} + \cdots \\ \Rightarrow {\left( {1 + \frac{1}{x}} \right)^x} = 1 + 1 + \frac{{{x^2}\left( {1 – \frac{1}{x}} \right)}}{{2!}}\frac{1}{{{x^2}}} + \frac{{{x^3}\left( {1 – \frac{1}{x}} \right)\left( {1 – \frac{2}{x}} \right)}}{{3!}}\frac{1}{{{x^3}}} + \cdots \\ \Rightarrow {\left( {1 + \frac{1}{x}} \right)^x} = 1 + 1 + \frac{1}{{2!}}\left( {1 – \frac{1}{x}} \right) + \frac{1}{{3!}}\left( {1 – \frac{1}{x}} \right)\left( {1 – \frac{2}{x}} \right) + \cdots \\ \end{gathered}$

Taking the limit as $x \to \infty$ for both sides, we get
$\begin{gathered} \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } \left[ {1 + 1 + \frac{1}{{2!}}\left( {1 – \frac{1}{x}} \right) + \frac{1}{{3!}}\left( {1 – \frac{1}{x}} \right)\left( {1 – \frac{2}{x}} \right) + \cdots } \right] \\ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = 1 + 1 + \frac{1}{{2!}}\mathop {\lim }\limits_{x \to \infty } \left( {1 – \frac{1}{x}} \right) + \frac{1}{{3!}}\mathop {\lim }\limits_{x \to \infty } \left( {1 – \frac{1}{x}} \right)\left( {1 – \frac{2}{x}} \right) + \cdots \\ \end{gathered}$

Applying limits we have
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = 1 + 1 + \frac{1}{{2!}}\left( {1 – \frac{1}{\infty }} \right) + \frac{1}{{3!}}\left( {1 – \frac{1}{\infty }} \right)\left( {1 – \frac{2}{\infty }} \right) + \cdots$

As we know that $\frac{1}{\infty } = 0$, we have
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = 1 + 1 + \frac{1}{{2!}}\left( {1 – 0} \right) + \frac{1}{{3!}}\left( {1 – 0} \right)\left( {1 – 0} \right) + \cdots$
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = 1 + 1 + \frac{1}{{2!}} + \frac{1}{{3!}} + \cdots \,\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

As we know that the series ${e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \cdots$,

putting $x = 1$ in the above series, we have
$e = 1 + 1 + \frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{4!}} + \cdots$

Using this value in equation (i), we have
$\Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = e$