# How do I us the Limit definition of derivative on f(x)=e^x?

Aug 6, 2014

The limit definition of the derivative is:

$\frac{d}{\mathrm{dx}} f \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

Now, since our function $f \left(x\right) = {e}^{x}$, we will substitute:

$\frac{d}{\mathrm{dx}} \left[{e}^{x}\right] = {\lim}_{h \to 0} \frac{{e}^{x + h} - {e}^{x}}{h}$

At first, it may be unclear as to how we will evaluate this limit. We will first rewrite it a bit, using a basic exponent law:

$\frac{d}{\mathrm{dx}} \left[{e}^{x}\right] = {\lim}_{h \to 0} \frac{{e}^{x} \cdot {e}^{h} - {e}^{x}}{h}$

And now, we will factor the ${e}^{x}$:

$\frac{d}{\mathrm{dx}} \left[{e}^{x}\right] = {\lim}_{h \to 0} \frac{{e}^{x} \left({e}^{h} - 1\right)}{h}$

It might not be obvious, but using the constant law of limits we can actually treat ${e}^{x}$ as a constant here and pull it out of the limit as a multiplier:

$\frac{d}{\mathrm{dx}} \left[{e}^{x}\right] = {e}^{x} \cdot {\lim}_{h \to 0} \frac{{e}^{h} - 1}{h}$

And now, the entire thing has been simplified a great deal. The tricky part is figuring out this last limit.

Since it's easier, we will attempt to evaluate the limit graphically. So let's take a look at a graph of the function $y = \frac{{e}^{x} - 1}{x}$ and see what happens when $x \to 0$:

The "hole" at $x = 0$ is caused by a division by zero. Thus, the function is undefined at $x = 0$. However, the function is well-defined everywhere else, even at values extremely close to zero. And, when $x$ gets extremely close to zero, we can see that $y$ appears to be getting closer to $1$:

$\frac{{e}^{0.1} - 1}{0.1} \approx 1.0517$

$\frac{{e}^{0.01} - 1}{0.01} \approx 1.0050$

$\frac{{e}^{0.001} - 1}{0.001} \approx 1.0005$

And, we can observe this same trend when approaching from the negative side:

$\frac{{e}^{-} 0.1 - 1}{-} 0.1 \approx 0.9516$

$\frac{{e}^{-} 0.01 - 1}{-} 0.01 \approx 0.9950$

$\frac{{e}^{-} 0.001 - 1}{-} 0.001 \approx 0.9995$

So, we can say with reasonable certainty that ${\lim}_{h \to 0} \frac{{e}^{h} - 1}{h} = 1$.

Granted, one shouldn't assume that they will get the correct answer from evaluating a limit graphically. So, since I like certainty, and since there is a way to evaluate the above limit algebraically, I will explain the alternate method:

${\lim}_{h \to 0} \frac{{e}^{h} - 1}{h}$

Now, there are actually a few ways to define $e$ itself as a limit. One of them is

$e = {\lim}_{h \to 0} {\left(1 + h\right)}^{\frac{1}{h}}$

Since our previous limit also has the variable $h$ approaching zero, we can actually substitute the definition of $e$.

${\lim}_{h \to 0} \frac{{\left({\left(1 + h\right)}^{\frac{1}{h}}\right)}^{h} - 1}{h}$

Simplifying the inside gives:

${\lim}_{h \to 0} \frac{1 + h - 1}{h}$

This further simplifies to:

${\lim}_{h \to 0} \frac{h}{h}$

We can easily see that this limit evaluates to $1$.

So now that we know what this limit is, we can look back at our definition for the derivative of ${e}^{x}$.

$\frac{d}{\mathrm{dx}} \left[{e}^{x}\right] = {e}^{x} \cdot {\lim}_{h \to 0} \frac{{e}^{h} - 1}{h}$

$= {e}^{x} \cdot 1$
$= {e}^{x}$