A Few Examples of Limit Proofs
Prove lim (7x − 4) = 10
x→2
SCRATCH WORK
First, we need to find a way of relating |x − 2| < δ and |(7x − 4) − 10| < ². We will use algebraic manipulation to get
this relationship. Remember that the whole point of this manipulation is to find a δ in terms of ² so that if |x − 2| < δ
is true for some x then it forces |(7x − 4) − 10| < ² to be true for that x.
So we will start with the ² term and manipulate it until we find the delta term in there.
So |(7x − 4) − 10| < ² ⇔ |7x − 4 − 10| < ² ⇔ |7x − 14| < ² ⇔ |7(x − 2)| < ² ⇔ |7| · |x − 2| < ² ⇔ 7|x − 2| < ²
Now 7|x − 2| < ² ⇔ |x − 2| <
²
7
Now with the above string of equivalences, we can see that if we let δ = 7² , then we have found a δ that will fulfill the
definition because if |x − 2| < δ then |x − 2| <
²
7
⇔ 7|x − 2| < ² ⇔ |7x − 14| < ² ⇔ |7x − 4 − 10| < ² ⇔ |(7x − 4) − 10| < ²
So we have that for any ² given to us if we let δ =
²
7
and |x − 2| < δ then this will force |(7x − 4) − 10| < ² thus
showing that the definition is fulfilled. Now let’s write the proof.
PROOF
Let ² > 0 be given. Let δ = 7² .
So if 0 < |x − 2| < δ then |x − 2| <
²
7
⇒ 7|x − 2| < ² ⇒ |7x − 14| < ² ⇒ |7x − 4 − 10| < ² ⇒ |(7x − 4) − 10| < ²
Therefore we have shown that lim (7x − 4) = 10. Done
x→2
By proving the limit, all we have really done is show that the definition is fulfilled.
Let’s do a few more. This time I will leave out a lot of the work shown above.
Prove lim (− 23 x + 4) = 0
x→6
SCRATCH WORK
Again we need a way to relate |x − 6| < δ and |(− 32 x + 4) − 0| < ²
Let’s start with the ² term and see if we can manipulate it down to be something related to the δ term.
|(− 23 x + 4) − 0| < ² ⇔ | − 23 x + 4| < ² ⇔ | − 23 (x − 6)| < ² ⇔ | − 32 | · |x − 6| < ² ⇔ 23 |x − 6| < ² ⇔ |x − 6| <
Therefore if we let δ =
3²
2
then if |x − 6| < δ then this forces |x − 6| <
3²
2
3²
2
⇔ 23 |x − 6| < ² ⇔ | − 32 | · |x − 6| < ² ⇔
| − 32 (x − 6)| < ² ⇔ | − 32 x + 4| < ² ⇔ | − 23 x + 4 − 0| < ² ⇔ |(− 32 x + 4) − 0| < ²
So for our choice of δ we can show that the definition is fulfilled.
Just a quick note on this one. You will be tempted to let δ = − 3²
2 which doesn’t make sense because in the definition
both δ and ² must be positive and if ² is positive then if we let δ = − 3²
2 then δ would be negative - which is very bad.
Now, let’s do the proof.
PROOF
Let ² > 0 be given. Let δ =
3²
2 .
So if 0 < |x − 6| < δ then |x − 6| <
3²
2
⇒ 23 |x − 6| < ² ⇒ | − 32 ||x − 6| < ² ⇒ | − 32 (x − 6)| < ² ⇒ | − 23 x + 4| < ² ⇒
| − 32 x + 4 − 0| < ² ⇒ |(− 32 x + 4) − 0| < ²
Therefore we have shown that lim (− 23 x + 4) = 0. Done
x→6
Make sure that you understand every implication (⇒) given in the above proof. Make sure you know why one
statement leads to the next statement.
Now we will do a proof where we need to restrict δ.
Prove lim (x2 − 3x + 1) = −1
x→2
SCRATCH WORK
So we need a way of relating |x − 2| < δ and |(x2 − 3x + 1) − (−1)| < ²
Let’s start with the ² term.
|(x2 − 3x + 1) − (−1)| < ² ⇔ |x2 − 3x + 1 + 1| < ² ⇔ |x2 − 3x + 2| < ² ⇔ |(x − 1)(x − 2)| < ² ⇔ |x − 1||x − 2| < ²
Now we run into a slight problem because we’ve got the δ term just like before but it’s not multiplied by a constant,
it’s multiplied by some function of x. The problem is that we can’t just divide by |x − 1| to get that |x − 2| <
²
|x−1|
=δ
because δ cannot be a function of x, only of ². So to fix this, we will restrict δ and by restricting δ we will actually be
restricting x because we are only concerned with when |x − 2| < δ.
Let’s choose some initial value to use as a restriction for δ. We will start with the value of 1. So let’s start by saying
δ < 1. If δ < 1 then |x − 2| < δ < 1 ⇒ |x − 2| < 1 ⇒ −1 < x − 2 < 1 ⇒ 1 < x < 3. Now with this domain of values for
x, what is the largest that |x − 1| can be? We get an upper bound of 2 meaning that |x − 1| < 2 if x is between 1 and 3.
So let’s look at where we left off with our ² term. We had gotten to |x − 1||x − 2| < ². Now with our restriction on
δ (i.e. δ < 1) this gives us a restriction on x (i.e. 1 < x < 3) which in turn gives us an upper bound for |x − 1| namely
|x − 1| < 2.
Therefore if 2|x − 2| < ² then |x − 1||x − 2| < ² because |x − 1||x − 2| < 2|x − 2| < ². So if we can get that 2|x − 2| < ²
then certainly |x − 1||x − 2| < ². So with 2|x − 2| < ² then |x − 2| <
²
2
which we will set δ equal to. So we will let δ = 2² .
Now let’s do the proof.
PROOF
Let ² > 0 be given. Let δ = min{ 2² , 1}.
Consider that if 0 < |x − 2| < δ < 1 then −1 < x − 2 < 1 ⇒ 1 < x < 3 ⇒ |x − 1| < 2.
Now if 0 < |x − 2| < δ ⇒ |x − 2| <
²
2
⇒ 2|x − 2| < ² ⇒ |x − 1||x − 2| < 2|x − 1| < ² (Make sure you know why this is
true.) ⇒ |x − 1||x − 2| < ² ⇒ |(x − 1)(x − 2)| < ² ⇒ |x2 − 3x + 2| < ² ⇒ |x2 − 3x + 1 + 1| < ² ⇒ |(x2 − 3x + 1) − (−1)| < ²
Therefore we have shown that lim (x2 − 3x + 1) = −1. Done
x→2
A few tips and tricks.
• 95% of the time, you should start with your ² term and algebraically manipulate it until you get something of the
form |g(x)| · |x − c| < ² where g(x) is some function of x.
• Once you find the function g(x) from the above hint, restrict δ (which forces a restriction on x) so that you can get
an upper bound for |g(x)|.
• Use this upper bound for |g(x)| to find δ. So if we algebraically get to |g(x)||x−c| < ² and |g(x)| < B for some number
B then if we can make B|x − c| < ² then this will force |g(x)||x − c| to be less than ² because |g(x)||x − c| < B|x − c| < ²
I’ve included a few more examples of some proofs. I haven’t put the scratch work in. Make sure that you can do the
scratch work to get the proof that I have written down.
Prove lim (2x + 3) = 1
x→−1
PROOF
Let ² > 0 be given. Let δ = 2² .
So 0 < |x−(−1)| < δ ⇒ |x+1| <
²
2
⇒ 2|x+1| < ² ⇒ |2||x+1| < ² ⇒ |2(x+1)| < ² ⇒ |2x+2| < ² ⇒ |(2x+3)−1| < ².
Therefore we have shown that lim (2x + 3) = 1. Done
x→−1
Prove lim (−3x + 1) = 7
x→−2
PROOF
Let ² > 0 be given. Let δ = 3² .
So 0 < |x − (−2)| < δ ⇒ |x + 2| <
²
3
⇒ 3|x + 2| < ² ⇒ | − 3||x + 2| < ² ⇒ | − 3(x + 2)| < ² ⇒ | − 3x − 6| < ² ⇒
|(−3x + 1) − 7| < ².
Therefore we have shown that lim (−3x + 1) = 7. Done
x→−2
Prove lim (x2 − 3x + 2) = 6
x→4
PROOF
Let ² > 0 be given. Let δ = min{ 6² , 1}.
Consider that if 0 < |x − 4| < δ < 1 then −1 < x − 4 < 1 ⇒ 3 < x < 5 ⇒ |x + 1| < 6.
Now 0 < |x − 4| < δ ⇒ |x − 4| <
²
6
⇒ 6|x − 4| < ² ⇒ |x + 1||x − 4| < 6|x − 4| < ² ⇒ |x + 1||x − 4| < ² ⇒
|(x + 1)(x − 4)| < ² ⇒ |x2 − 3x − 4| < ² ⇒ |x2 − 3x + 2 − 6| < ² ⇒ |(x2 − 3x + 2) − 6| < ²
Therefore we have shown that lim (x2 − 3x + 2) = 6. Done
x→4
4x
= −4
x→2 x − 4
Prove lim
PROOF
Let ² > 0 be given. Let δ = min{ 8² , 1}.
Consider that if 0 < |x − 2| < δ < 1 then −1 < x − 2 < 1 ⇒ 1 < x < 3 ⇒
8
< 8.
|x − 4|
8
8
Now 0 < |x − 2| < δ ⇒ |x − 2| < 8² ⇒ 8|x − 2| < ² ⇒
|x − 2| < 8|x − 2| < ² ⇒
· |x − 2| < ² ⇒
|x
−
4|
|x
−
¯
¯
¯
¯
¯
¯
¯4|
¯
¯
¯
¯
¯
¯
¯
¯
¯
|8x − 16|
8x − 16 ¯
4x + 4x − 16 ¯
4x + 4(x − 4) ¯
4x
8|x − 2|
¯
¯
¯
¯
<²⇒
<²⇒¯
<²⇒¯
<²⇒¯
<²⇒¯
+ 4¯¯ < ² ⇒
¯
¯
¯
|x − 4| ¶
x−4
x−4
x−4
x−4
¯µ
¯|x − 4|
¯ 4x
¯
¯
¯
¯ x − 4 − (−4)¯ < ²
4x
Therefore we have shown that lim
= −4. Done
x→2 x − 4