Limit as x -> 3 of (x^2 - 9)/(x - 3) using L'Hôpital's rule Step by Step

Step by step

The question

Question: find the limit as x → 3 of (x² − 9)/(x − 3).

lim x → 3 x² − 9x − 3 =

1 Put the number in

Put 3 in place of x.

The top is 0, and the bottom is 0.

top: 3² − 9 = 0

bottom: 3 − 3 = 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 2x.

The derivative of the bottom is 1.

The new limit is the same.

lim x → 3 x² − 9x − 3 = lim x → 3 2x1

3 Put the number in again

Put 3 in again.

Now the bottom is not 0.

So the limit is 6.

= 2(3)1 = 61 = 6

4 Check

Let's check with a calculator, with x very close to 3.

x = 2.99 gives about 5.99.

x = 2.99: x² − 9x − 3 ≈ 5.99

And x = 3.01 gives about 6.01.

Both are close to 6.

Right!

x = 3.01: x² − 9x − 3 ≈ 6.01

Let's look back.

First: put the number in.

Then: use L'Hôpital's rule.

Then: put the number in again.

Last: check.

Answer