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

The limit definition of the derivative is:

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

Now, since our function ##f(x) = e^x##, we will substitute:

##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 ##e^x##:

##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 ##e^x## as a constant here and pull it out of the limit as a multiplier:

##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 ##y = (e^x – 1)/x## and see what happens when ##x->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##:

##(e^0.1 – 1)/0.1 approx 1.0517 ##

##(e^0.01 – 1)/0.01 approx 1.0050 ##

##(e^0.001 – 1)/0.001 approx 1.0005 ##

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

##(e^-0.1 – 1)/-0.1 approx 0.9516 ##

##(e^-0.01 – 1)/-0.01 approx 0.9950 ##

##(e^-0.001 – 1)/-0.001 approx 0.9995 ##

So, we can say with reasonable certainty that ##lim_(h->0) (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->0) (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->0) (1 + h)^(1/h)##

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

##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 ##1##.

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

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

## = e^x * 1## ##= e^x##