Limit as x -> 0 of (e^(2x) - 1)/sin(x) Step by Step

Step by step

The question

Question: find the limit as x → 0 of (e²ˣ − 1) over sin x.

lim x → 0 e^2x − 1sin x =

1 Put the number in

Put 0 in place of x.

The top is 0, and the bottom is 0.

top: e⁰ − 1 = 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 2e²ˣ.

The derivative of the bottom is cos x.

The new limit is the same.

lim x → 0 e^2x − 1sin x = lim x → 0 2e^2xcos x

3 Put the number in again

Put 0 in again.

Now the bottom is not 0.

So the limit is 2.

= 2e⁰cos 0 = 21 = 2

4 Check

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

x = −0.01 gives about 1.9802.

x = −0.01: e^2x − 1sin x ≈ 1.9802

And x = 0.01 gives about 2.0202.

Both are close to 2.

Right!

x = 0.01: e^2x − 1sin x ≈ 2.0202

Let's look back.

First: put the number in.

Then: use L'Hôpital's rule.

Then: put the number in again.

Last: check.

Answer