The function the limit of (1+x/n)^n as n->infinity

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Now let us examine the function

$\displaystyle f(x)=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{x}{n}} \right)}^{n}}$

We know it is defined for at least one value of x, and it is quite easy to see that it will also converge to a number between 1 and e for x between 0 and 1.

We can virtually copy-paste the binomial expansion from the previous page to get

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{x}{n}} \right)}^{n}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {1+\left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right){{{\left( {\frac{x}{n}} \right)}}^{1}}+\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){{{\left( {\frac{x}{n}} \right)}}^{2}}+\left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right){{{\left( {\frac{x}{n}} \right)}}^{3}}+...} \right)$

this will give us

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{x}{n}} \right)}^{n}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {1+n\cdot \frac{x}{n}+\frac{{n\left( {n-1} \right)}}{{2\cdot 1}}\frac{{{{x}^{2}}}}{{{{n}^{2}}}}+\frac{{n\left( {n-1} \right)\left( {n-2} \right)}}{{3\cdot 2\cdot 1}}\frac{{{{x}^{3}}}}{{{{n}^{3}}}}+...} \right)$

And in the same way as in the previous page, we will get

$\displaystyle f(x)=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{x}{n}} \right)}^{n}}=1+\frac{x}{{1!}}+\frac{{{{x}^{2}}}}{{2!}}+\frac{{{{x}^{3}}}}{{3!}}+\frac{{{{x}^{4}}}}{{4!}}+...$

Next, what happens if we take the derivative of this function? We get

$\displaystyle f\acute{\ }(x)=0+\frac{1}{{1!}}+\frac{{2x}}{{2!}}+\frac{{3{{x}^{2}}}}{{3!}}+\frac{{4{{x}^{3}}}}{{4!}}+...=1+\frac{x}{{1!}}+\frac{{{{x}^{2}}}}{{2!}}+\frac{{{{x}^{3}}}}{{3!}}+...$

I.e., the same function as we started with.

Up a level : Euler's number e
Previous page : The limit of (1+1/n)^n as n->infinity
Next page : The function e^x

Last modified: Dec 27, 2023 @ 22:39