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

##### 1 Answer

The limit definition of the derivative is:

#d/dx f(x) = lim_(h->0) (f(x+h) - f(x))/h#

Now, since our function

#d/dx[e^x] = lim_(h->0) (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:

#d/dx[e^x] = lim_(h->0) (e^(x) * e^h - e^x)/h#

And now, we will factor the

#d/dx[e^x] = lim_(h->0) (e^x (e^h - 1))/h#

It might not be obvious, but using the constant law of limits we can actually treat

#d/dx[e^x] = e^x * lim_(h->0) (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

The "hole" at *appears* to be getting closer to

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

So, we can say with reasonable certainty that

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->0) (e^h - 1)/h#

Now, there are actually a few ways to define

#e = lim_(h->0) (1 + h)^(1/h)#

Since our previous limit also has the variable

#lim_(h->0) (((1 + h)^(1/h))^h - 1)/h#

Simplifying the inside gives:

#lim_(h->0) (1 + h - 1)/h#

This further simplifies to:

#lim_(h->0) h/h#

We can easily see that this limit evaluates to

So now that we know what this limit is, we can look back at our definition for the derivative of

#d/dx[e^x] = e^x * lim_(h->0) (e^h - 1)/h#

# = e^x * 1#

#= e^x#