Limits:
At the end of our previous lesson, we were left with a situation where the gradient of the tangent
of a curve at point 𝑥, could be approximated by:
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
𝑚=
∆𝑥
where we wanted ∆𝑥 to become almost 0, without actually being 0.
The larger the value of ∆𝑥 is, the worse the approximation of the gradient will be.
The smaller ∆𝑥 is, the better the approximation of the gradient will be.
BUT if ∆𝑥 = 0 then the gradient between the two points will be undefined.
This shows us that we need a mathematical tool that allows us to get ∆𝑥 as close as possible to
zero, without being zero. Fortunately, limits can solve this problem for us.
What are limits?
The value of a function, 𝑓(𝑥), as its 𝑥-value “gets close” to some point of value 𝑎 is known as “the
limit of the function, 𝑓(𝑥) as 𝑥 tends towards 𝑎”.
Mathematically this is written as:
lim 𝑓(𝑥)
𝑥→𝑎
𝑥 does not have to actually reach a value of 𝑎, rather it must approach a value of 𝑎.
Example:
Consider the function 𝒇(𝒙) = 𝒙𝟐 + 𝟏, determine the limit of the function, 𝒇(𝒙) as 𝒙 tends towards
2.
First let’s use a graph to visualise what is being asked of us:
By looking at the graph we know that when
𝑥 = 2 then 𝑦 = (2¿2 + 1 = 5.
But what happens to the function value
immediately to the left or to the right of
𝑥 = 2?
from the left:
𝒚 𝑥 < 2 starting further away and getting
→? closer in value to 2.
𝑥 𝑓(𝑥)
0 ( )
𝑓0 =1
1 𝑓(1) = 2
1,5 𝑓(1,5) = 3,25
1,75 𝑓(1,75) = 4,0625
𝒙→𝟐
1,8 𝑓(1,8) = 4,24
1,9 𝑓(1,9) = 4,61
1,99 𝑓(1,99) = 4,9601
1,999 𝑓(1,999) = 4,996001
1,9999 𝑓(1,9999) = 4,99960001
1,99999 𝑓(1,99999) = 4,99996
1
, From the right:
𝑥 > 2 starting further away and getting closer in value to 2.
𝑥 𝑓(𝑥)
4 𝑓(4) = 17
3 𝑓(3) = 10
2,5 𝑓(2,5) = 7,25
2,2 𝑓(2,2) = 5,84
2,1 𝑓(2,1) = 5,41
2,01 𝑓(2,01) = 5,0401
2,001 𝑓(2,001) = 5,004001
2,0001 𝑓(2,0001) = 5,00040001
2,00001 𝑓(2,00001) = 5,0000400001
2,000001 𝑓(2,000001) = 5,000004000001
In this way we can see that whether we start from the left of 2 or from the right of 2, as our 𝑥 value
approaches 2 our function value approaches 5.
In this instance we see that
lim 𝑓(𝑥) = 𝑓(2)
𝑥→2
Limits by substitution
The easiest way to find a limit is to substitute 𝑥 for the desired value of 𝑥 and see if you obtain a
result, which will then be the limit. This is a neat trick to determine which 𝑦-value is being
approached.
What
NOTE:doSubstitution
we mean about a function
will ONLY work being continuous
if the function at a point? at that particular 𝑥-value. If
is continuous
the function is not continuous at that particular 𝑥-value an attempt to substitute may give you
Aan
function can limit
undefined be either:
even if the limit exists.
Continuous – it has no gaps, sudden changes in value or vertical asymptotes, its domain is
𝑥 ∈ ℝ.
2
At the end of our previous lesson, we were left with a situation where the gradient of the tangent
of a curve at point 𝑥, could be approximated by:
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
𝑚=
∆𝑥
where we wanted ∆𝑥 to become almost 0, without actually being 0.
The larger the value of ∆𝑥 is, the worse the approximation of the gradient will be.
The smaller ∆𝑥 is, the better the approximation of the gradient will be.
BUT if ∆𝑥 = 0 then the gradient between the two points will be undefined.
This shows us that we need a mathematical tool that allows us to get ∆𝑥 as close as possible to
zero, without being zero. Fortunately, limits can solve this problem for us.
What are limits?
The value of a function, 𝑓(𝑥), as its 𝑥-value “gets close” to some point of value 𝑎 is known as “the
limit of the function, 𝑓(𝑥) as 𝑥 tends towards 𝑎”.
Mathematically this is written as:
lim 𝑓(𝑥)
𝑥→𝑎
𝑥 does not have to actually reach a value of 𝑎, rather it must approach a value of 𝑎.
Example:
Consider the function 𝒇(𝒙) = 𝒙𝟐 + 𝟏, determine the limit of the function, 𝒇(𝒙) as 𝒙 tends towards
2.
First let’s use a graph to visualise what is being asked of us:
By looking at the graph we know that when
𝑥 = 2 then 𝑦 = (2¿2 + 1 = 5.
But what happens to the function value
immediately to the left or to the right of
𝑥 = 2?
from the left:
𝒚 𝑥 < 2 starting further away and getting
→? closer in value to 2.
𝑥 𝑓(𝑥)
0 ( )
𝑓0 =1
1 𝑓(1) = 2
1,5 𝑓(1,5) = 3,25
1,75 𝑓(1,75) = 4,0625
𝒙→𝟐
1,8 𝑓(1,8) = 4,24
1,9 𝑓(1,9) = 4,61
1,99 𝑓(1,99) = 4,9601
1,999 𝑓(1,999) = 4,996001
1,9999 𝑓(1,9999) = 4,99960001
1,99999 𝑓(1,99999) = 4,99996
1
, From the right:
𝑥 > 2 starting further away and getting closer in value to 2.
𝑥 𝑓(𝑥)
4 𝑓(4) = 17
3 𝑓(3) = 10
2,5 𝑓(2,5) = 7,25
2,2 𝑓(2,2) = 5,84
2,1 𝑓(2,1) = 5,41
2,01 𝑓(2,01) = 5,0401
2,001 𝑓(2,001) = 5,004001
2,0001 𝑓(2,0001) = 5,00040001
2,00001 𝑓(2,00001) = 5,0000400001
2,000001 𝑓(2,000001) = 5,000004000001
In this way we can see that whether we start from the left of 2 or from the right of 2, as our 𝑥 value
approaches 2 our function value approaches 5.
In this instance we see that
lim 𝑓(𝑥) = 𝑓(2)
𝑥→2
Limits by substitution
The easiest way to find a limit is to substitute 𝑥 for the desired value of 𝑥 and see if you obtain a
result, which will then be the limit. This is a neat trick to determine which 𝑦-value is being
approached.
What
NOTE:doSubstitution
we mean about a function
will ONLY work being continuous
if the function at a point? at that particular 𝑥-value. If
is continuous
the function is not continuous at that particular 𝑥-value an attempt to substitute may give you
Aan
function can limit
undefined be either:
even if the limit exists.
Continuous – it has no gaps, sudden changes in value or vertical asymptotes, its domain is
𝑥 ∈ ℝ.
2
