Step by step
The question
Question: find the limit as x → 9 of (x − 9) over (√x − 3).
1 Put the number in
Put 9 in place of x.
The top is 0, and the bottom is 0.
The square root of 9 is 3.
3 minus 3 is 0.
0/0 has no value.
It tells us to work more.
2 Multiply by the conjugate
The conjugate of √x − 3 is √x + 3: the same two terms, with the sign swapped.
Multiply the top and the bottom by the conjugate.
Something over itself is 1.
So the value does not change.
Careful!
Both the top and the bottom!
One alone would change the value.
3 Multiply out
Remember: a minus b, times a plus b, is a squared minus b squared.
The middle terms cancel.
At the bottom: (√x − 3) times (√x + 3).
That is the root squared, minus 3 squared.
The top we leave as it is.
A square root squared is just what is inside.
So it is x − 9.
Now (x − 9) is on the top and on the bottom.
4 Cancel
x gets close to 9, but it is never 9.
So x − 9 is not 0.
Something over itself is 1.
So we cross out (x − 9) on the top and the bottom.
Write what is left: √x + 3.
5 Put the number in again
Now the 0 is gone.
Put 9 in again.
The square root of 9 is 3.
So the limit is 6.
6 Check
Let's check with numbers close to 9.
A calculator gives 5.9983 at x = 8.99.
And 6.0017 at x = 9.01.
Both are close to 6.
Right!
Let's look back.
First: put the number in.
Then: multiply by the conjugate.
Then: multiply out.
Then: cancel.
Then: put the number in again.
Last: check.