Steply

Limit as x -> 0 of (1 - cos(2x))/x^2

Step by step, with the reason for every step · L'Hôpital's rule · Calculus 1

Step by step

The question

Question: find the limit as x → 0 of (1 − cos(2x)) over x².

lim x → 0 1 − cos(2x)x2 =

1 Put the number in

Put 0 in place of x.

The top is 0, and the bottom is 0.

top: 1 − cos 0 = 0
bottom: 02 = 0

0/0 has no value.

It tells us nothing yet.

This is where L'Hôpital's rule helps.

form 00

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.

lim x → a f(x)g(x) = lim x → a f′(x)g′(x)

Careful!

This is not the quotient rule.

The top gets its own derivative, and the bottom gets its own derivative.

Careful: not the quotient rule: top and bottom each alone

The derivative of the top is 2 sin(2x).

The derivative of the bottom is 2x.

The new limit is the same.

lim x → 0 1 − cos(2x)x2 = lim x → 0 2 sin(2x)2x

3 Use it again

Put 0 in place of x.

The top is 0, and the bottom is 0.

top: 2sin 0 = 0
bottom: 2(0) = 0

0/0 again!

So we use the rule again.

form 00

The derivative of the top is 4 cos(2x).

The derivative of the bottom is 2.

The new limit is the same.

= lim x → 0 4 cos(2x)2

4 Put the number in again

Put 0 in again.

Now the bottom is not 0.

So the limit is 2.

= 4cos 02 = 42 = 2

5 Check

Let's check with a calculator in radians, with x very close to 0.

x = −0.01 gives about 1.9999.

x = −0.01: 1 − cos(2x)x2 ≈ 1.9999

And x = 0.01 gives about 1.9999.

Both are close to 2.

Right!

x = 0.01: 1 − cos(2x)x2 ≈ 1.9999

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.

Last: check.

Answer
= 4cos 02 = 42 = 2

Your turn

Try this one yourself, then check:

lim x → 0 3x2−cos x + 1
Show the answer
6

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