Worked examples
The question
Question: find the limit as x → 3 of (x² + 2x − 1).
lim x → 3 (x2 + 2x − 1) =
1 Put the number in
This is a polynomial.
Its graph has no gaps and no jumps.
So the limit is just its value at x = 3.
no gaps, no jumps → just put it in
Put 3 in place of x.
= 32 + 2(3) − 1
= 9 + 6 − 1
So the limit is 14.
= 14
2 Check
Let's check with numbers close to 3.
x = 2.99 gives about 13.9201.
x = 2.99: f(x) ≈ 13.9201
x = 3.01 gives about 14.0801.
Both are close to 14.
Right!
x = 3.01: f(x) ≈ 14.0801
Let's look back.
First: put the number in.
Last: check.
▶ Watch it solved
The question
Question: find the limit as x → 2 of (x² − 4)/(x − 2).
lim x → 2 x2 − 4x − 2 =
1 Put the number in
Put 2 in place of x.
The top is 0, and the bottom is 0.
22 − 42 − 2 = 00
0/0 has no value.
It does not tell us the limit.
It tells us to work more.
00: no value yet. Work more!
Careful!
0/0 is not 0, and it is not 1.
Some piece of the top and the bottom is 0 at 2.
We find it.
Careful: not 00 = 0 ✗
2 Factor
When 0/0 comes out, (x − 2) hides in the top and the bottom.
Factor them.
lim x → 2 x2 − 4x − 2 = lim x → 2 (x − 2)(x + 2)(x − 2)
x² − 4 is a difference of two squares: x² minus 2 squared.
So it is (x − 2)(x + 2).
a² − b² = (a − b)(a + b), b = 2
3 Cancel
x gets close to 2, but it is never 2.
So x − 2 is not 0.
x ≠ 2, so x − 2 ≠ 0
A number over itself is 1.
So we cross out (x − 2) on the top and the bottom.
Write what is left: x + 2.
= lim x → 2 (x + 2)
4 Put the number in again
Now x − 2 is gone, so the bottom is not 0.
Put 2 in again.
= (2 + 2)
So the limit is 4.
= 4
5 Check
Let's check with numbers close to 2.
x = 1.99 gives about 3.99.
x = 1.99: f(x) ≈ 3.99
x = 2.01 gives about 4.01.
Both are close to 4.
Right!
x = 2.01: f(x) ≈ 4.01
Let's look back.
First: put the number in.
Then: factor.
Then: cancel.
Then: put the number in again.
Last: check.
▶ Watch it solved
The question
Question: find the limit as x → 3 of (x² − 9)/(x² − x − 6).
lim x → 3 x2 − 9x2 − x − 6 =
1 Put the number in
Put 3 in place of x.
The top is 0, and the bottom is 0.
32 − 932 − 3 − 6 = 00
0/0 has no value.
It does not tell us the limit.
It tells us to work more.
00: no value yet. Work more!
Careful!
0/0 is not 0, and it is not 1.
Some piece of the top and the bottom is 0 at 3.
We find it.
Careful: not 00 = 0 ✗
2 Factor
When 0/0 comes out, (x − 3) hides in the top and the bottom.
Factor them.
lim x → 3 x2 − 9x2 − x − 6 = lim x → 3 (x − 3)(x + 3)(x − 3)(x + 2)
x² − 9 is a difference of two squares: x² minus 3 squared.
So it is (x − 3)(x + 3).
a² − b² = (a − b)(a + b), b = 3
We need two numbers that multiply to −6 and add to −1: 2 and −3.
2 × (−3) = −6, 2 + (−3) = −1
3 Cancel
x gets close to 3, but it is never 3.
So x − 3 is not 0.
x ≠ 3, so x − 3 ≠ 0
A number over itself is 1.
So we cross out (x − 3) on the top and the bottom.
Write what is left: (x + 3)/(x + 2).
= lim x → 3 (x + 3)(x + 2)
4 Put the number in again
Now x − 3 is gone, so the bottom is not 0.
Put 3 in again.
= (3 + 3)(3 + 2)
So the limit is 6/5.
= 65
5 Check
Let's check with numbers close to 3.
x = 2.99 gives about 1.2004.
x = 2.99: f(x) ≈ 1.2004
x = 3.01 gives about 1.1996.
Both are close to 6/5.
Right!
x = 3.01: f(x) ≈ 1.1996
Let's look back.
First: put the number in.
Then: factor.
Then: cancel.
Then: put the number in again.
Last: check.
▶ Watch it solved
