Step by step
The question
Question: find the limit as x → 0 of (1 − cos(2x)) over x².
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 2 sin(2x).
The derivative of the bottom is 2x.
The new limit is the same.
3 Use it again
Put 0 in place of x.
The top is 0, and the bottom is 0.
0/0 again!
So we use the rule again.
The derivative of the top is 4 cos(2x).
The derivative of the bottom is 2.
The new limit is the same.
4 Put the number in again
Put 0 in again.
Now the bottom is not 0.
So the limit is 2.
5 Check
Let's check with a calculator in radians, with x very close to 0.
x = −0.01 gives about 1.9999.
And x = 0.01 gives about 1.9999.
Both are close to 2.
Right!
6 Pick the choice
Our answer is 2.
That is choice C.
7 Why not the others
A is 0.
Wrong: stopped after one use of the rule: 2sin(2x)/(2x) is still 0/0.
B is 1/2.
Wrong: lost the 2s from the chain rule: cos(2x) gives −2sin(2x).
D is 4.
Wrong: did not divide by 2: the second derivative of x² is 2.
Let's look back.
First: put the number in.
Then: use L'Hôpital's rule.
Then: use it again.
Then: put the number in again.
Then: check.
Then: pick the choice.
Last: why not the others.