Steply

Limit as x -> 0 of sin(5x)/sin(2x)

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 sin(5x) over (sin(2x)).

lim x → 0 sin(5x)sin(2x) =

1 Put the number in

Put 0 in place of x.

The top is 0, and the bottom is 0.

top: sin 0 = 0
bottom: sin 0 = 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 5 cos(5x).

The derivative of the bottom is 2 cos(2x).

The new limit is the same.

lim x → 0 sin(5x)sin(2x) = lim x → 0 5 cos(5x)2 cos(2x)

3 Put the number in again

Put 0 in again.

Now the bottom is not 0.

So the limit is 5/2.

= 5cos 02cos 0 = 52 = 52

4 Check

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

x = −0.01 gives about 2.4991.

x = −0.01: sin(5x)sin(2x) ≈ 2.4991

And x = 0.01 gives about 2.4991.

Both are close to 5/2.

Right!

x = 0.01: sin(5x)sin(2x) ≈ 2.4991

Let's look back.

First: put the number in.

Then: use L'Hôpital's rule.

Then: put the number in again.

Last: check.

Answer
= 5cos 02cos 0 = 52 = 52

Your turn

Try this one yourself, then check:

lim x → 0 sin(2x)sin(4x)
Show the answer
12

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