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 have 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 by 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. And as 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}}=\underset{{x\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{x}{n}} \right)}^{n}}

\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}}

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