# The function e^x

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So, what kind of function is this f(x) that we defined in the previous page?

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

We will find this out in two different ways. To start with, we can see that f(1)=1.

Next, let us multiply f by e. This will give us

$\displaystyle \begin{array}{l}f(x)\cdot e=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{x}{n}} \right)}^{n}}\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{1}{n}} \right)}^{n}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{x}{n}} \right)}^{n}}{{\left( {1+\frac{1}{n}} \right)}^{n}}\\\quad \quad \quad =\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {\left( {1+\frac{x}{n}} \right)\left( {1+\frac{1}{n}} \right)} \right)}^{n}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{x}{n}+\frac{1}{n}+\frac{x}{{{{n}^{2}}}}} \right)}^{n}}\end{array}$

Next, we can make the three last terms having a common denominator

$\displaystyle \begin{array}{l}f(x)\cdot e=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{{x+1+\frac{x}{n}}}{n}} \right)}^{n}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{{x+1+\frac{x}{n}}}{n}} \right)}^{n}}\\\quad \quad \quad =\underset{{n\to \infty }}{\mathop{{\lim }}}\,f\left( {x+1+\frac{x}{n}} \right)=f\left( {x+1} \right)\end{array}$

So, every time we multiply by e we will increase the input parameter by one, and f(1)=1. Hmm, what kind of function are dealing with here? Ok, it is of course an exponential function with the base e, i.e.,

$\displaystyle f(x)={{e}^{x}}$

Let us try to show this in a second way. We could start with looking at the expected result. Say we have

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

We could do that last step since n is independent of x.  Then we do a binomial expansion of this, or rather, a generalized binomial expansion to get

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

Just copying the reasoning from the first page in this section. An as in there we can see that most of those terms will go toward 0. This gives us

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

And our function is thus ex.

We have now established a few important facts.

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

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

and

$\displaystyle {{\left( {{{e}^{x}}} \right)}^{\prime }}={{e}^{x}}$

Up a level : Euler's number e
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