Worked examples
The question
Question: decide if f(x) = x² − 3 is continuous at x = 2.
f(x) = x2 − 3
Continuous at x = 2 means three things are true.
One: f(2) has a value.
Two: the limit there exists.
Three: they are equal.
1. f(2) has a value
2. the limit at 2 exists
3. limit = f(2)
1 Does f have a value?
Put 2 in place of x.
We get 1.
f(2) = 22 − 3 = 4 − 3 = 1
Careful!
Having a value is not enough.
The graph must also arrive at that height from both sides.
Careful: a value at 2 is not enough
Check 1 is true: f(2) is 1.
1. f(2) = 1
2 Does the limit exist?
A polynomial has no gaps and no jumps.
So its limit at 2 is just its value: 1.
polynomial: no gaps, no jumps
So the limit is 1.
lim x → 2 f(x) = 1
Check 2 is true: the limit exists.
2. limit = 1
3 Are they equal?
The limit is 1, and f(2) is 1.
They are equal.
Check 3 is true.
3. limit = 1 = f(2)
All three checks are true.
So f is continuous at x = 2.
= f is continuous at x = 2
4 Check
Let's check on the graph.
The graph is one piece with no gaps.
It passes through (2, 1) without lifting the pen.
Right!
Let's look back.
First: does f have a value?.
Then: does the limit exist?.
Then: are they equal?.
Last: check.
Answer = f is continuous at x = 2
▶ Watch it solved
The question
Question: decide if this f(x) is continuous at x = 1.
f(x) = x + 2 if x < 1
= 3x if x ≥ 1
Continuous at x = 1 means three things are true.
One: f(1) has a value.
Two: the limit there exists.
Three: they are equal.
1. f(1) has a value
2. the limit at 1 exists
3. limit = f(1)
1 Does f have a value?
Which piece holds x = 1?
The one with x ≥ 1.
So f(1) = 3.
f(1) = 3(1) = 3
Careful!
Having a value is not enough.
The graph must also arrive at that height from both sides.
Careful: a value at 1 is not enough
Check 1 is true: f(1) is 3.
1. f(1) = 3
2 Does the limit exist?
From the left, we use x + 2.
We get 3.
lim x → 1⁻ f(x) = 1 + 2 = 3
From the right, we use 3x.
We get 3.
lim x → 1⁺ f(x) = 3(1) = 3
Both sides give 3.
Check 2 is true: the limit exists.
2. left = right = 3
3 Are they equal?
The limit is 3, and f(1) is 3.
They are equal.
Check 3 is true.
3. limit = 3 = f(1)
All three checks are true.
So f is continuous at x = 1.
= f is continuous at x = 1
4 Check
Let's check on the graph.
The first piece comes in from the left.
The second piece goes on from the same point.
No lifting the pen.
Right!
Let's look back.
First: does f have a value?.
Then: does the limit exist?.
Then: are they equal?.
Last: check.
Answer = f is continuous at x = 1
▶ Watch it solved
The question
Question: decide if this f(x) is continuous at x = 2.
f(x) = x2 − 4x − 2 if x ≠ 2
= 4 if x = 2
Continuous at x = 2 means three things are true.
One: f(2) has a value.
Two: the limit there exists.
Three: they are equal.
1. f(2) has a value
2. the limit at 2 exists
3. limit = f(2)
1 Does f have a value?
Which piece holds x = 2?
The one with x = 2.
So f(2) = 4.
f(2) = 4
Careful!
Having a value is not enough.
The graph must also arrive at that height from both sides.
Careful: a value at 2 is not enough
Check 1 is true: f(2) is 4.
1. f(2) = 4
2 Does the limit exist?
Near 2, but not at 2, f(x) is (x² − 4)/(x − 2).
Factor the top and the bottom.
lim x → 2 f(x) = lim x → 2 (x − 2)(x + 2)(x − 2)
x is never 2 here, so x − 2 is not 0.
Cross it out on the top and the bottom.
Put 2 in what is left.
The limit is 4.
= lim x → 2 (x + 2) = 4
Check 2 is true: the limit exists.
2. limit = 4
3 Are they equal?
The limit is 4, and f(2) is 4.
They are equal.
Check 3 is true.
3. limit = 4 = f(2)
All three checks are true.
So f is continuous at x = 2.
= f is continuous at x = 2
4 Check
Let's check on the graph.
The graph comes to height 4 from both sides, but there is a hole at x = 2.
f(2) = 4 fills the hole.
No lifting the pen.
Right!
Let's look back.
First: does f have a value?.
Then: does the limit exist?.
Then: are they equal?.
Last: check.
Answer = f is continuous at x = 2
▶ Watch it solved
