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