Worked examples
The question
Question: find the limit as x → 2 of this f(x): x + 1 if x < 2, and 5 − x if x ≥ 2.
f(x) = x + 1 if x < 2
= 5 − x if x ≥ 2
1 From the left
From the left means x is a bit less than 2.
There, f(x) is x + 1.
lim x → 2⁻ f(x) = lim x → 2⁻ (x + 1)
So we use the first piece.
x < 2: piece 1
A polynomial piece has no gaps, so we just put 2 in.
We get 3.
= 2 + 1 = 3
2 From the right
From the right means x is a bit more than 2.
There, f(x) is 5 − x.
lim x → 2⁺ f(x) = lim x → 2⁺ (5 − x)
So we use the second piece.
x > 2: piece 2
A polynomial piece has no gaps, so we just put 2 in.
We get 3.
= 5 − 2 = 3
3 Compare the sides
From the left we get 3.
From the right we get 3.
left: 3 right: 3
Both sides give the same number.
So the limit is 3.
lim x → 2 f(x) = 3
Careful!
One side is not enough.
A limit needs both sides to give the same number.
Careful: both sides must give the same number
4 Check
Let's check on the graph.
The first piece comes in from the left.
It reaches height 3.
The second piece starts at height 3.
They meet: no jump.
At x = 2 both pieces arrive at 3.
Right!
Let's look back.
First: from the left.
Then: from the right.
Then: compare the sides.
Last: check.
Answer Careful: both sides must give the same number
▶ Watch it solved
The question
Question: find the limit as x → 2 of this f(x): x² if x < 2, and x + 2 if x ≥ 2.
f(x) = x2 if x < 2
= x + 2 if x ≥ 2
1 From the left
From the left means x is a bit less than 2.
There, f(x) is x².
lim x → 2⁻ f(x) = lim x → 2⁻ x2
So we use the first piece.
x < 2: piece 1
A polynomial piece has no gaps, so we just put 2 in.
We get 4.
= 22 = 4
2 From the right
From the right means x is a bit more than 2.
There, f(x) is x + 2.
lim x → 2⁺ f(x) = lim x → 2⁺ (x + 2)
So we use the second piece.
x > 2: piece 2
A polynomial piece has no gaps, so we just put 2 in.
We get 4.
= 2 + 2 = 4
3 Compare the sides
From the left we get 4.
From the right we get 4.
left: 4 right: 4
Both sides give the same number.
So the limit is 4.
lim x → 2 f(x) = 4
Careful!
One side is not enough.
A limit needs both sides to give the same number.
Careful: both sides must give the same number
4 Check
Let's check on the graph.
The first piece comes in from the left.
It reaches height 4.
The second piece starts at height 4.
They meet: no jump.
At x = 2 both pieces arrive at 4.
Right!
Let's look back.
First: from the left.
Then: from the right.
Then: compare the sides.
Last: check.
Answer Careful: both sides must give the same number
▶ Watch it solved
The question
Question: find the limit as x → 1 of 2/(x − 1)².
lim x → 1 2(x − 1)2 =
1 Put the number in
Put 1 in place of x.
The top is 2, and the bottom is 0.
2(1 − 1)2 = 20
A number over 0 has no value.
And it is not 0/0, so there is nothing to cancel.
a number over 0: no value
not 0 over 0: nothing cancels
Careful!
2/0 is not 0.
We look at each side instead.
Careful: not 20 = 0 ✗
2 From the left
Numbers just below 1: 0.9, 0.99, 0.999.
The bottom is a tiny positive number.
So the values get bigger and bigger.
200
20000
2000000
They go up without end.
We write ∞: infinity.
from the left: f(x) → ∞
3 From the right
Numbers just above 1: 1.1, 1.01, 1.001.
The bottom is a tiny positive number.
So the values get bigger and bigger.
200
20000
2000000
They go up without end.
We write ∞: infinity.
from the right: f(x) → ∞
4 Compare the sides
From the left we get ∞.
From the right we get ∞.
left: ∞ right: ∞
Both sides go up without end, the same way.
So the limit is ∞: f(x) grows without end.
lim x → 1 2(x − 1)2 = ∞
5 Check
Let's check on the graph.
The dashed line is x = 1.
The graph never touches it.
Left of it, the graph shoots up.
Right of it, the graph shoots up.
Right!
Let's look back.
First: put the number in.
Then: from the left.
Then: from the right.
Then: compare the sides.
Last: check.
Answer lim x → 1 2(x − 1)2 = ∞
▶ Watch it solved
