Step by step
The question
Question: find the limit as x → 0 of (e³ˣ − 1) over x.
There are four choices.
First we solve it, the normal way.
Then we pick.
1 Put the number in
Put 0 in place of x.
The top is 0, and the bottom is 0.
0/0 has no value.
It tells us nothing yet.
This is where L'Hôpital's rule helps.
2 Use L'Hôpital's rule
The rule: when we get 0/0, or infinity over infinity, we may take the derivative of the top, and the derivative of the bottom.
The new limit is the same.
Careful!
This is not the quotient rule.
The top gets its own derivative, and the bottom gets its own derivative.
The derivative of the top is 3e³ˣ.
The derivative of the bottom is 1.
The new limit is the same.
3 Put the number in again
Put 0 in again.
Now the bottom is not 0.
So the limit is 3.
4 Check
Let's check with a calculator, with x very close to 0.
x = −0.01 gives about 2.9554.
And x = 0.01 gives about 3.0455.
Both are close to 3.
Right!
5 Pick the choice
Our answer is 3.
That is choice D.
6 Why not the others
A is 0.
Wrong: 0/0 is not 0.
B is 1.
Wrong: lost × 3 from the derivative of e³ˣ.
C is 1/3.
Wrong: turned it upside down.
Let's look back.
First: put the number in.
Then: use L'Hôpital's rule.
Then: put the number in again.
Then: check.
Then: pick the choice.
Last: why not the others.