Limits
Before giving a rigorous definition of the notion of a limit, we will give the following statement
as an intuitive definition of a limit.
Let 𝑓(𝑥) be a real valued function defined on 𝑅. If the value of 𝑓(𝑥) gets closer and closer to
the value 𝐿 as 𝑥 gets closer and closer to 𝑎, then we say that the limit of 𝑓(𝑥) as 𝑥 approaches
𝑎 is 𝐿.
Example 1
Let 𝑓(𝑥) = 𝑥 3 ∀𝑥
As 𝑓(2) = 8, any student would expect the limit of 𝑓(𝑥) as 𝑥 approaches 2 to be 8. Let us
consider some of the 𝑥 values which gets closer and closer to 2.
𝑓(2.1) = 9.261
𝑓(2.01) = 8.1206
𝑓(2.001) = 8.012
𝑓(2.0001) = 8.0012
It is clear that the value of 𝑓(𝑥) gets closer and closer to 8. But, it also gets closer and closer
to 8.001. So, how do we know that the limit is not 8.001?
Our candidate for the limit was 8 because 𝑓(2) = 8. but the following example would convince
you that it was not necessary.
Example 2
𝑥3 ∀𝑥 ≠ 2
𝑓(𝑥) = {
0 𝑖𝑓 𝑥 = 2
Again,
𝑓(2.1) = 9.261
𝑓(2.01) = 8.1206
𝑓(2.001) = 8.012
𝑓(2.0001) = 8.0012
𝑓(2.00001) = 8.00012
We will get the same sequence of function values even if we define
𝑓(𝑥) = 𝑥 3 ∀ 𝑥 ≠ 2. So, what happens precisely at the point at which we want to find the limit
seems to be unimportant.
Now, we need to understand the concept of a neighbourhood.
Definition 1
Let 𝑎 ∈ 𝑅. A neighbourhood of 𝑎 is an open interval (𝑎 − 𝛿, 𝑎 + 𝛿) for any 𝛿 > 0. A deleted
neighbourhood of 𝑎 is (𝑎 − 𝛿, 𝑎 + 𝛿) \{𝑎} .
From what we have seen above, when taking the limit of a function as 𝑥 approaches 𝑎, deleted
neighbourhoods of 𝑎 are more important than the neighbourhoods of 𝑎. If the limit of 𝑓(𝑥) as
1
, 𝑥 approaches 𝑎 is 𝐿, we can write that fact as lim 𝑓(𝑥) = 𝐿. Now, we have a feeling that when
𝑥→𝑎
lim 𝑓(𝑥) = 𝐿, we can make 𝑓(𝑥) values as close to 𝐿 as we wish for all 𝑥 values sufficiently
𝑥→𝑎
close to 𝑎 except 𝑎 itself.
Note that making 𝑓(𝑥) arbitrarily close to 𝐿 is the same as making
|𝑓(𝑥) − 𝐿| arbitrarily small. With that observation, we are ready for the formal definition of a
limit.
Definition 2
Let 𝑓(𝑥) be a real valued function defined on an open interval containing 𝑎 possibly (but not
necessarily) excluding 𝑎 itself. We say that the limit of 𝑓(𝑥) as 𝑥 approaches 𝑎 is 𝐿 if ∀𝜀 > 0,
∃ 𝛿 > 0 such that
|𝑓(𝑥) − 𝐿| < 𝜀 ∀𝑥 ∈ (𝑎 − 𝛿, 𝑎 + 𝛿) ∖ {𝑎}
Note that an equivalent statement to this would be ∀𝜀 > 0, ∃ 𝛿 > 0 such that 0 < |𝑥 − 𝑎| <
𝛿 ⟹ |𝑓(𝑥) − 𝐿| < 𝜀
Now, we will show how to prove what we have seen in the example 2 using this definition.
For all the 𝑥 values such that 𝑥 ≠ 2,
𝑓(𝑥) − 8 = 𝑥 3 − 8 = (𝑥 − 2)(𝑥 2 + 2𝑥 + 4)
∴ |𝑓(𝑥) − 8| = |𝑥 − 2||𝑥 2 + 2𝑥 + 4|
≤ |𝑥 − 2|{|𝑥|2 + 2|𝑥| + 4}
Let 𝜀 < 0. Our eventual goal is getting |𝑓(𝑥) − 𝐿| < 𝜀.
We do that in the following manner.
0 < |𝑥 − 2| < 1 ⟹ |𝑥| − 2 < 1 ⇒ |𝑥 2 | + 2|𝑥| + 4 < 9 + 2(3) + 4 ⇒ |𝑥|2 + 2|𝑥| + 4
< 19 ⇒ |𝑥 − 2|{|𝑥|2 + 2|𝑥| + 4} < 19|𝑥 − 2| ⇒ |𝑓(𝑥) − 8| < 19|𝑥 − 2|
Define
𝜀
𝛿 = 𝑀𝑖𝑛 {1, } .
19
𝜀
(The student should think why we have there. )
19
Now,
0 < |𝑥 − 2| < 𝛿 ⟹ |𝑓(𝑥) − 8| < 19|𝑥 − 2|
𝜀
⟹ |𝑓(𝑥) − 8| < 19 ( ) = 𝜀
19
∴ By the definition of a limit, lim 𝑓(𝑥) = 8.
𝑥→𝑎
Theorem 1
For any 𝑎 ∈ 𝑅, lim 𝑥 𝑛 = 𝑎𝑛 where 𝑛 is a positive integer.
𝑥→𝑎
Proof:
Use the factorization
𝑥 𝑛 − 𝑎𝑛 = (𝑥 − 𝑎)(𝑥 𝑛−1 + 𝑥 𝑛−2 𝑎 + …….. +𝑥𝑎𝑛−2 + 𝑎𝑛−1 )
2
Before giving a rigorous definition of the notion of a limit, we will give the following statement
as an intuitive definition of a limit.
Let 𝑓(𝑥) be a real valued function defined on 𝑅. If the value of 𝑓(𝑥) gets closer and closer to
the value 𝐿 as 𝑥 gets closer and closer to 𝑎, then we say that the limit of 𝑓(𝑥) as 𝑥 approaches
𝑎 is 𝐿.
Example 1
Let 𝑓(𝑥) = 𝑥 3 ∀𝑥
As 𝑓(2) = 8, any student would expect the limit of 𝑓(𝑥) as 𝑥 approaches 2 to be 8. Let us
consider some of the 𝑥 values which gets closer and closer to 2.
𝑓(2.1) = 9.261
𝑓(2.01) = 8.1206
𝑓(2.001) = 8.012
𝑓(2.0001) = 8.0012
It is clear that the value of 𝑓(𝑥) gets closer and closer to 8. But, it also gets closer and closer
to 8.001. So, how do we know that the limit is not 8.001?
Our candidate for the limit was 8 because 𝑓(2) = 8. but the following example would convince
you that it was not necessary.
Example 2
𝑥3 ∀𝑥 ≠ 2
𝑓(𝑥) = {
0 𝑖𝑓 𝑥 = 2
Again,
𝑓(2.1) = 9.261
𝑓(2.01) = 8.1206
𝑓(2.001) = 8.012
𝑓(2.0001) = 8.0012
𝑓(2.00001) = 8.00012
We will get the same sequence of function values even if we define
𝑓(𝑥) = 𝑥 3 ∀ 𝑥 ≠ 2. So, what happens precisely at the point at which we want to find the limit
seems to be unimportant.
Now, we need to understand the concept of a neighbourhood.
Definition 1
Let 𝑎 ∈ 𝑅. A neighbourhood of 𝑎 is an open interval (𝑎 − 𝛿, 𝑎 + 𝛿) for any 𝛿 > 0. A deleted
neighbourhood of 𝑎 is (𝑎 − 𝛿, 𝑎 + 𝛿) \{𝑎} .
From what we have seen above, when taking the limit of a function as 𝑥 approaches 𝑎, deleted
neighbourhoods of 𝑎 are more important than the neighbourhoods of 𝑎. If the limit of 𝑓(𝑥) as
1
, 𝑥 approaches 𝑎 is 𝐿, we can write that fact as lim 𝑓(𝑥) = 𝐿. Now, we have a feeling that when
𝑥→𝑎
lim 𝑓(𝑥) = 𝐿, we can make 𝑓(𝑥) values as close to 𝐿 as we wish for all 𝑥 values sufficiently
𝑥→𝑎
close to 𝑎 except 𝑎 itself.
Note that making 𝑓(𝑥) arbitrarily close to 𝐿 is the same as making
|𝑓(𝑥) − 𝐿| arbitrarily small. With that observation, we are ready for the formal definition of a
limit.
Definition 2
Let 𝑓(𝑥) be a real valued function defined on an open interval containing 𝑎 possibly (but not
necessarily) excluding 𝑎 itself. We say that the limit of 𝑓(𝑥) as 𝑥 approaches 𝑎 is 𝐿 if ∀𝜀 > 0,
∃ 𝛿 > 0 such that
|𝑓(𝑥) − 𝐿| < 𝜀 ∀𝑥 ∈ (𝑎 − 𝛿, 𝑎 + 𝛿) ∖ {𝑎}
Note that an equivalent statement to this would be ∀𝜀 > 0, ∃ 𝛿 > 0 such that 0 < |𝑥 − 𝑎| <
𝛿 ⟹ |𝑓(𝑥) − 𝐿| < 𝜀
Now, we will show how to prove what we have seen in the example 2 using this definition.
For all the 𝑥 values such that 𝑥 ≠ 2,
𝑓(𝑥) − 8 = 𝑥 3 − 8 = (𝑥 − 2)(𝑥 2 + 2𝑥 + 4)
∴ |𝑓(𝑥) − 8| = |𝑥 − 2||𝑥 2 + 2𝑥 + 4|
≤ |𝑥 − 2|{|𝑥|2 + 2|𝑥| + 4}
Let 𝜀 < 0. Our eventual goal is getting |𝑓(𝑥) − 𝐿| < 𝜀.
We do that in the following manner.
0 < |𝑥 − 2| < 1 ⟹ |𝑥| − 2 < 1 ⇒ |𝑥 2 | + 2|𝑥| + 4 < 9 + 2(3) + 4 ⇒ |𝑥|2 + 2|𝑥| + 4
< 19 ⇒ |𝑥 − 2|{|𝑥|2 + 2|𝑥| + 4} < 19|𝑥 − 2| ⇒ |𝑓(𝑥) − 8| < 19|𝑥 − 2|
Define
𝜀
𝛿 = 𝑀𝑖𝑛 {1, } .
19
𝜀
(The student should think why we have there. )
19
Now,
0 < |𝑥 − 2| < 𝛿 ⟹ |𝑓(𝑥) − 8| < 19|𝑥 − 2|
𝜀
⟹ |𝑓(𝑥) − 8| < 19 ( ) = 𝜀
19
∴ By the definition of a limit, lim 𝑓(𝑥) = 8.
𝑥→𝑎
Theorem 1
For any 𝑎 ∈ 𝑅, lim 𝑥 𝑛 = 𝑎𝑛 where 𝑛 is a positive integer.
𝑥→𝑎
Proof:
Use the factorization
𝑥 𝑛 − 𝑎𝑛 = (𝑥 − 𝑎)(𝑥 𝑛−1 + 𝑥 𝑛−2 𝑎 + …….. +𝑥𝑎𝑛−2 + 𝑎𝑛−1 )
2
