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
\[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = e\]
Qin
August 15 @ 2:32 pm
Hi:
I have a further question that how to prove e=1+x+x^2/2!+… without using taylor expansion.Because I think taylor expansion is based on derivative of e^x. And in order to derive the derivative of e^x, we may need a lot of some other derivatives or limits which finally resorts to lim(1+1’x)^x=e . And they cause a proof loop.so we may find a another approach to e’s proof
Aditya Ravindran
January 26 @ 2:41 pm
e was defined as this limit. This limit comes from the compound interest problem.