functions, domain,range, continuity

1. The word Calculus comes from Latin meaning "small stone",
Because it is like understanding something by looking at small pieces
Differential Calculus cuts something into small pieces to find how it changes.
Real time Application of Calculus
 Space flight engineers frequently use calculus when planning for long missions. To
launch a space rocket/satellite, they must consider the different orbiting velocities
of the Earth and the planet the rocket is targeted for, as well as other gravitational
influences like the sun and the moon.
Differential Calculus

2. • An Architect Engineer uses integration in determining the amount of the necessary materials to
construct curved shape constructions (e.g. dome over a sports arena) and also to measure the
weight of that structure. Calculus is used to improve the architecture not only of buildings but
also of important infrastructures such as bridges.

3. • In robotics , the mobile robots drive around, with a velocity (the
derivative of position) and an acceleration (the derivative of velocity).
Robotic arms move, each joint has an angular velocity and acceleration.

4. Some Basic Terminologies

6. Example: y = x2
• x is an Independent Variable
• y is the Dependent Variable
When x=2 , y = 4
When x=3 , y =9
etc
Example: h = 2w + d
• w is an Independent Variable
• d is an Independent Variable
• h is the Dependent Variable

9. (𝟑, 𝟔) - Open - It contains all numbers between 3 & 6
(excluding 3 &6)
[𝟑, 𝟔) - Half closed & half open – It contains all numbers
between 3 & 6 (including 3 & excluding 6)
(𝟑, 𝟔] - Half open & half closed – It contains all numbers
between 3 & 6 (excluding 3 & including 6)
[𝟑, 𝟔] - Closed – It contains all numbers between 3 & 6
(including 3 & 6)

10. Cartesian Product
The Cartesian product between two sets A and B is the set of all possible
ordered pairs(a, b) with first element a from A and second element b from B
(i.e) A X B = {(a, b): a ∈ A, b ∈ B}.
For example,
Consider two non-empty sets
𝐴 = 𝑎1, 𝑎2, 𝑎3 and 𝐵 = {𝑏1, 𝑏2, 𝑏3}
Cartesian product A X B
= {(a1,b1), (a1,b2), (a1,b3), ( a2,b1), (a2,b2),(a2,b3), (a3,b1), (a3,b2), (a3,b3)}.

11. Another Example,
Consider the two non-empty sets 𝐴 = 𝑥, 𝑦, 𝑧 & 𝐵 = {1,2,3}
Cartesian product
𝐴 × 𝐵 = {(𝑥, 1), (𝑥, 2), (𝑥, ), ( 𝑦, 1), (𝑦, 2), (𝑦, 3), (𝑧, 1), (𝑧, 2), (𝑧, 3)}

12. Note ∶
1. 𝐴 𝑋 𝐵 ≠ 𝐵 𝑋 𝐴
2. If either A or B is empty set, then A X B = Φ
3. Two ordered pairs are equal, if and only if the corresponding first elements are
equal and the second elements are also equal.
4. If there are m elements in A and n elements in B, then there will be mn
elements in A × B, i.e., if n(A) = m and n(B) = n, then n(A × B) = mn.
5. If A and B are non-empty sets and either A or B is an infinite set,
then so is A × B.

13. Relation
A relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B. The subset is
derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
Remarks
1. The second element is called the image of the first element.
2. The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of
the relation R.
3. The set of all second elements in a relation R from a set A to a set B is called the range of the relation R. The
whole set B is called the codomain of the relation R.
4. range ⊆ codomain.
5. The total number of relations that can be defined from a set A to a set B is the number of possible subsets of
A × B. If n(A ) = p and n(B) = q, then n (A × B) = pq and the total number of relations is 𝟐 𝒑𝒒
.

14. Examples
Let A = {1, 2, 3,...,14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, where x, y є A}.
Write down its domain, codomain and range.
Given 𝟑𝒙 − 𝒚 = 𝟎
⟹ 𝒚 = 𝟑𝒙
𝑾𝒉𝒆𝒏 𝒙 = 𝟏 , 𝒚 = 𝟑 𝟏 = 𝟑 ∈ 𝑨
𝑾𝒉𝒆𝒏 𝒙 = 𝟐 , 𝒚 = 𝟑 𝟐 = 𝟔 ∈ 𝑨
𝑾𝒉𝒆𝒏 𝒙 = 𝟑 , 𝒚 = 𝟑 𝟑 = 𝟗 ∈ 𝑨
𝑾𝒉𝒆𝒏 𝒙 = 𝟒 , 𝒚 = 𝟑 𝟒 = 𝟏𝟐 ∈ 𝑨
𝑾𝒉𝒆𝒏 𝒙 = 𝟓 , 𝒚 = 𝟑 𝟓 = 𝟏𝟓 ∉ 𝑨
⟹ R = {(x, y) : 3x – y = 0, where x, y є A}= {(1,3),(2,6),(3,9),(4,12)}
Domain = {1, 2, 3, 4}
Codomain = A = {1, 2, 3, ..., 14}.
Range = {3, 6, 9, 12}

15. 𝑥 𝑦 = 3𝑥
Range
Or Image
Domain Codomain
1
2
3
4
5
6
7
8
9
10
11
12
13
14
14

16. Function
Function is like a machine which has input and output.
The machine has no wheels or movement and it will not destroy the input.
Every input has a unique output.
A function relates every input with exactly one output.

17. A function is a relation which describes that there should be only one output
for each input.
A function is a special kind of relation (a set of ordered pairs), which follows
a rule.
i.e every x-value should be associated with only one y-value is called a
function.
The set of all functions is a subset of the set of all relations
A function with domain A and codomain B , denoted by f : A→B , is
relation between the elements of A and B satisfying the properties:
for all a ∈ A, there is a unique b ∈ B such that f(a) =b .
Here ‘b’ is called the image of ‘a’ under f

18. Example
Consider A = { 1,2,3} and B = {a, b, c, d}
Here f: A → B is a function
A = {1, 2, 3} is called domain of f
B = {a, b, c, d} is called co-domain of f
Here {a, b, c} is called range or image of f
And {1, 2, 3} is called pre-image of {a, b, c}
A f B
1
2
3
3
a
b
c
d

19. Consider the following
A B
1
2
a
b
This is not a function, since the element 1 of A had two images a and b
Example

20. Graph of a function
The graph of a function f is the set of all points in the plane of the form (x, f(x)).
The graph of a function is the set of all points whose co-ordinates (x, y) satisfy the function y = f(x).
(i.e) the graph of f to be the graph of the equation y = f(x).

21. Vertical line test
The vertical line test can be used to determine whether a graph represents a
function.
A vertical line includes all points with a particular x value.
The y value of a point where a vertical line intersects a graph represents an
output for that input x value.
If we can draw any vertical line that intersects a graph more than once, then the
graph does not define a function because that x value has more than one output.
A function has only one output value for each input value.

22. Examples

23. Consider 𝑓(𝑥) = 2 Consider 𝑓(𝑥) = 𝑥
Identity FunctionConstant Function
https://www.desmos.com/calculator/iybkkybyf3 https://www.desmos.com/calculator/mtkqm3hutb

24. Consider 𝒇(𝒙) = 𝒙
Quadratic Function/Parabola
Consider 𝒇(𝒙) = 𝒙𝟐
Absolute value
Function/Modulus
https://www.desmos.com/calculator/obyhumnazb
https://www.desmos.com/calculator/vz0eg4ycly

25. Consider 𝒇(𝒙) = 𝒙𝟑 Reciprocal Function
Consider 𝒇( 𝒙) =
𝟏
𝒙
https://www.desmos.com/calculator/3yybwkbkeo https://www.desmos.com/calculator/kltvtf0wid

26. The domain of a function is the complete set of possible values of the independent variable (x)
Domain and Range of a Function
Domain
What can go into a function is called the Domain
Codomain
What may possibly come out of a function is called the Codomain
Range
What actually comes out of a function is called the Range
The range of a function is the complete set of all possible resulting values of the dependent variable (y),
after we have substituted the domain.

27. Note:
Rational function:
A rational function is a function of the form 𝑓(𝑥) =
𝑝 𝑥
𝑞 𝑥
, where 𝑝(𝑥)
and 𝑞(𝑥) are polynomials and 𝑞(𝑥) ≠ 0
Examples : 𝑓 𝑥 =
𝑥2−𝑥+1
𝑥2+𝑥+1
, 𝑦 =
𝑥2+𝑥−2
𝑥2−𝑥−2
Radical function:
A radical function is a form 𝑥 = 𝑝(𝑥) , where 𝑝(𝑥) is a polynomial.
Examples: 𝑓(𝑥) = 16 − 𝑥2 , 𝑦 = 𝑥 , 𝑦 = 𝑥2 + 2

28. Methods to find domain of a function
For Rational Functions :-
The domain of a rational function consists of all the real numbers 𝑥 except
those for which the denominator is 0.
To find these 𝑥 values to be excluded from the domain of a rational function,
equate the denominator to zero and solve for.
For Radical Functions :-
Inside of a radical cannot be negative if we want real numbers only. So, the
inside of a radical has to be 0 or a positive number and no negative numbers.
To find the 𝑥 values to be excluded from the domain of a radical function, take
𝑝(𝑥) ≥ 0 and solve for x.

29. Note :
When finding the domain, remember:
1.The denominator (bottom) of a fraction cannot be zero.
(i.e) There are no zero values in the denominator (bottom) of a fraction
2. The number under a square root sign must be positive in this section
(i.e) There are no negative values under a square root sign

30. Methods to find range of a function
To find the range of a function 𝑓(𝑥)
1.Take 𝑦 = 𝑓(𝑥) and then solve the equation for 𝑥, write in the form 𝑥 = 𝑔(𝑦).
2. Find the domain of 𝑔(𝑦), and this will be the range of 𝑓(𝑥).

31. Problem 1.
Find the range and domain of 𝑓(𝑥) = 16 − 𝑥2
Solution:
Given that 𝑓(𝑥) = 16 − 𝑥2 , which is a radical function
To find domain:
Take 16 − 𝑥2 ≥ 0
⇒ 𝑥2 − 16 ≤ 0
⇒ (𝑥 − 4)(𝑥 + 4) ≤ 0
Possible cases :
𝐶𝑎𝑠𝑒 𝑖 . (−)(+) ≤ 0
𝐶𝑎𝑠𝑒 𝑖𝑖 . (+)(−) ≤ 0
𝑪𝒂𝒔𝒆 (𝒊). (𝑥 − 4) ≤ 0 & ( 𝑥 + 4 ) ≥ 0
⟹ 𝑥 ≤ 4 & 𝑥 ≥ −4
⟹ −4 ≤ 𝑥 ≤ 4
𝑪𝒂𝒔𝒆 𝒊𝒊 . (𝑥 − 4) > 0 & ( 𝑥 + 4 ) < 0
𝑥 > 4 & 𝑥 < −4 (not possible)
Hence the domain is −4 ≤ 𝑥 ≤ 4 ⇒ 𝑥 ∈ [−4,4]
To find the range:
Let y = f(x)
𝑦 = √(16 – 𝑥2
)
Squaring both sides
𝑦2
= 16 – 𝑥2
𝑥2
= 16 − 𝑦2
⇒ 𝑥 = 16 – 𝑦2
⇒ 16 – 𝑦2
≥ 0
⇒ (y - 4) ( y + 4 ) ≤ 0
⇒ -4 ≤ y ≤ 4 ⇒ y ∈ [-4,4].
But y = √(16 – 𝑥2
) ≥ 0 for all x ∈ [-4,4].
∴ Range is [0,4].

32. Problem 2.
Find the domain of y =
𝑥2+𝑥−2
𝑥2−𝑥−2
Solution:
Given that y =
𝑥2+𝑥−2
𝑥2−𝑥−2
, which is a rational
function
To find domain:
The domain is all numbers except the 𝐷𝑟 = 0
Take 𝑥2 − 𝑥 − 2 = 0
⇒ (𝑥 − 2)(𝑥 + 1) = 0
⟹ 𝑥 = 2 𝑜𝑟 𝑥 = −1
Hence the domain is all numbers except 2 & -1
Problem 3.
Find the range of 𝒇 𝒙 =
𝒙 𝟐−𝒙+𝟏
𝒙 𝟐+𝒙+𝟏
Solution:
Given that 𝑓 𝑥 =
𝑥2−𝑥+1
𝑥2+𝑥+1
which is a rational function
To find the range:
Let y =
𝑥2−𝑥+1
𝑥2+𝑥+1
𝑦 𝑥2 + 𝑥 + 1 = 𝑥2 − 𝑥 + 1
𝑦𝑥2 + 𝑦𝑥 + 𝑦 − 𝑥2 + 𝑥 − 1 = 0
𝑦 − 1 𝑥2 + 𝑦 + 1 𝑥 + 𝑦 − 1 = 0 ------- (1)

33. Since x є R, and (1) is quadratic in x, x do not have a complex
value.
Hence (1) has real roots, and 𝑦 + 1 2 − 4 𝑦 − 1 𝑦 − 1 ≥ 0
(W.k.t, for real roots 𝑏2
− 4𝑎𝑐 ≥ 0)
𝑖. 𝑒, 𝑦2
+ 2𝑦 + 1 − 4𝑦2
+ 8𝑦 − 4 ≥ 0
𝑖. 𝑒, −3𝑦2
+ 10𝑦 − 3 ≥ 0
3𝑦2
− 10𝑦 + 3 ≤ 0
3𝑦2 − 9𝑦 − 𝑦 + 3 ≤ 0
3𝑦(𝑦 − 3) − (𝑦 − 3) ≤ 0
(𝑦 − 3)(3𝑦 − 1) ≤ 0
Possible cases :
𝑪𝒂𝒔𝒆 𝒊 . (−)(+) ≤ 0
𝑪𝒂𝒔𝒆 𝒊𝒊 . (+)(−) ≤ 0
𝑪𝒂𝒔𝒆 (𝒊).
(𝑦 − 3) ≤ 0 𝑎𝑛𝑑 (3𝑦 − 1) ≥ 0
𝑦 ≤ 3 𝑎𝑛𝑑 3𝑦 ≥ 1
𝑦 ≤ 3 𝑎𝑛𝑑 𝑦 ≥ 1/3
𝑪𝒂𝒔𝒆 (𝒊𝒊)
(𝑦 − 3) ≥ 0 𝑎𝑛𝑑 (3𝑦 − 1) ≤ 0
𝑦 ≥ 3 𝑎𝑛𝑑 3𝑦 ≤ 1
𝑦 ≥ 3 𝑎𝑛𝑑 𝑦 ≤ 1/3 (not possible )
Hence 𝑦 ∈ [
1
3
, 3] ⇒ the range is [
1
3
, 3]

34. Even and Odd Function
Even function
A function 𝑓(𝑥) is said to be even if
𝑓 −𝑥 = 𝑓 𝑥 , for all 𝑥
Examples
1. 𝑓 𝑥 = 𝑥2
2. 𝑓 𝑥 = 𝑥
3. 𝑓 𝑥 = 𝑐𝑜𝑠𝑥
4.𝑓 𝑥 = 𝑥𝑠𝑖𝑛𝑥
Note:
Even functions are symmetric about
y- axis https://www.desmos.com/calculator/gcodlke1qv

35. Odd function
A function 𝑓(𝑥) is said to be odd if
𝑓(−𝑥) = − 𝑓(𝑥)
Examples
1.𝑓 𝑥 = 𝑥
2. 𝑓 𝑥 = 𝑠𝑖𝑛𝑥
3.𝑓 𝑥 = 𝑥𝑐𝑜𝑠𝑥
4. 𝑓 𝑥 = 𝑥3
Note:
Odd functions are symmetric about
the origin
(𝑖. 𝑒) Odd functions are symmetric
in opposite quadrants https://www.desmos.com/calculator/gcodlke1qv

36. Problem 1
Determine whether the function 𝒈 𝒙 = 𝟏 + 𝟑𝒙 𝟐
− 𝒙 𝟒
is even or odd
Solution:
Given 𝑔 𝑥 = 1 + 3𝑥2 − 𝑥4 ------ (1)
Put x = -x in (1)
𝑔 −𝑥 = 1 + 3(−𝑥)2
−(−𝑥)4
= 1 + 3𝑥2
− 𝑥4
= 𝑔(𝑥)
∴ 𝑔 −𝑥 = 𝑔 𝑥
Hence g(x) is an even function

37. Limits
Limits are all about approaching.
If the limit of the function 𝑓(𝑥) is ‘L’
as x approaches the point ‘a’,
then we mentioned this as lim
𝑥⟶𝑎
𝑓 𝑥 = 𝐿
For Example, consider f x =
𝑥2 − 1
𝑥 − 1
Let's work it out for 𝑥 ⟶ 1:
𝑓 𝑥 =
𝑥2 − 1
𝑥 − 1
=
12 − 1
1 − 1
=
1 − 1
1 − 1
=
0
0
it is "indeterminate",
so we need another way of answering this.
So let's try approaching 1 closer and closer:
𝑥 𝑓(𝑥) =
𝑥2 − 1
𝑥 − 1
0.5 1.50000
0.9 1.90000
0.99 1.99000
0.999 1.99900
0.9999 1.99990
0.99999 1.99999
...
...

38. It is clear that as 𝑥 approaches 1 , 𝑓(𝑥) approaches 2
The limit of 𝑓(𝑥) =
𝑥2−1
𝑥−1
as 𝑥 approaches 1 is 2
And it is written in symbols as: lim
𝑥⟶1
𝑓 𝑥 = lim
𝑥⟶1
𝑥2−1
𝑥−1
= 𝟐
So, in truth, we cannot say what the value at 𝑥 = 1 is.
But we can say that as we approach 1, the limit is 2.

39. Left & Right Hand Limits (One Side limits)
The Left –hand limit
The left-hand limit of 𝑓 (𝑥), as 𝑥 approaches 𝑎, equals 𝐿 can be written as lim
𝑥⟶𝑎−
𝑓 𝑥 = 𝐿
If we can make the value 𝑓 (𝑥) arbitrarily close to 𝐿, by taking 𝑥 to be sufficiently close to the left of 𝑎.

40. The right-hand limit
The right-hand limit of 𝑓 (𝑥), as 𝑥 approaches 𝑎, equals 𝐿 can be written as lim
𝑥⟶𝑎+
𝑓 𝑥 = 𝐿
If we can make the value 𝑓 (𝑥) arbitrarily close to L, by taking 𝑥 to be sufficiently close to the right of 𝑎.
Hence 𝐥𝐢𝐦
𝒙⟶𝒂
𝒇 𝒙 = 𝑳 if and only if 𝐥𝐢𝐦
𝒙⟶𝒂−
𝒇(𝒙) = 𝐥𝐢𝐦
𝒙⟶𝒂+
𝒇 𝒙 = 𝑳

41. When the limit of 𝑓(𝑥) is different from different sides (left & Right)
The limit does not exist at "a"

42. Problem 1
Find the left & right hand limits of
𝒇 𝒙 =
𝒙 𝟐
𝒊𝒇 𝒙 ≤ 𝟑
𝟐𝒙 𝒊𝒇 𝒙 > 𝟑
Solution:
Right Hand limit:
lim
𝑥→3+
𝑓 𝑥 = lim
𝑥→3+
2𝑥 = 2 3 = 6
Left Hand limit:
lim
𝑥→3−
𝑓 𝑥 = lim
𝑥→3−
𝑥2
= 32
= 9
Problem 2
Find the left & right hand limits of
𝒇 𝒙 =
𝒙 + 𝟏, 𝒊𝒇 𝒙 > 𝟎
𝒙 − 𝟏, 𝒊𝒇 𝒙 ≤ 𝟎
Solution:
Right Hand limit
lim
𝑥→0+
𝑓 𝑥 = lim
𝑥→0+
𝑥 + 1 = 1
Left Hand limit
lim
𝑥→0−
𝑓 𝑥 = lim
𝑥→0−
𝑥 − 1 = −1
Note:
To find the Left & Right hand limit at x = 1
R.H.L lim
𝑥→1+
𝑓 𝑥 = lim
𝑥→1+
𝑥 + 1 = 1 + 1 = 2
L.H.L lim
𝑥→1−
𝑓 𝑥 = lim
𝑥→1−
𝑥 + 1 = 1 + 1 = 2

43. Properties of limits
1. lim
𝑥→𝑎
𝑥 = 𝑎
2. lim
𝑥→𝑎
𝑥 𝑛 = 𝑎 𝑛
3. lim
𝑥→𝑎
𝑐 = 𝑐 , where c is a constant
4. If lim
𝑥→𝑎
𝑓(𝑥) = 𝐿, then lim
𝑥→𝑎
𝑐𝑓(𝑥) = 𝑐𝐿
5. If lim
𝑥→𝑎
𝑓(𝑥) = 𝐿, then lim
𝑥→𝑎
𝑓(𝑥) = 𝐿 , 𝐿 > 0
6. If lim
𝑥→𝑎
𝑓(𝑥) = 𝐿, then lim
𝑥→𝑎
𝑓 𝑥
𝑛
= 𝐿 𝑛
7. If lim
𝑥→𝑎
𝑓(𝑥) = 𝐿 and lim
𝑥→𝑎
𝑔(𝑥) = 𝑀, then
lim
𝑥→𝑎
(𝑓(𝑥) ± 𝑔(𝑥)) = lim
𝑥→𝑎
𝑓 𝑥 ± lim
𝑥→𝑎
𝑔 𝑥 = 𝐿 ± 𝑀
8. If lim
𝑥→𝑎
𝑓(𝑥) = 𝐿 and lim
𝑥→𝑎
𝑔(𝑥) = 𝑀, then
lim
𝑥→𝑎
(𝑓(𝑥) 𝑔(𝑥)) = lim
𝑥→𝑎
𝑓 𝑥 lim
𝑥→𝑎
𝑔 𝑥 = 𝐿𝑀
9. If lim
𝑥→𝑎
𝑓(𝑥) = 𝐿 and lim
𝑥→𝑎
𝑔(𝑥) = 𝑀, then
lim
𝑥→𝑎
𝑓 𝑥
𝑔 𝑥
=
lim
𝑥→𝑎
𝑓 𝑥
lim
𝑥→𝑎
𝑔 𝑥
=
𝐿
𝑀
, 𝑀 ≠ 0

44. Results
1. lim
𝑥→0
𝑒 𝑥−1
𝑥
= 1
2. lim
𝑥→0
𝑎 𝑥−1
𝑥
= log 𝑒 𝑎 , 𝑎 > 0
3. lim
𝑥→0
log(1+𝑥)
𝑥
= 1
4. lim
𝑥→∞
1 +
1
𝑥
𝑥
= 𝑒
5. lim
𝑥→0
1 + 𝑥
1
𝑥 = 𝑒
6. lim
𝑥→∞
1 +
𝑎
𝑥
𝑥
= 𝑒 𝑎
7. lim
𝑥→𝑎
𝑥 𝑛−𝑎 𝑛
𝑥−𝑎
= 𝑛𝑎 𝑛−1
8. lim
𝑥→0
(1+𝑥) 𝑛−1
𝑥
= 𝑛
9. lim
𝑥→∞
𝑒 𝑥 = ∞
10. lim
𝑥→∞
𝑒−𝑥
= ∞
11. lim
𝑥→0
1
𝑥
= 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡 (𝑙𝑒𝑓𝑡 ℎ𝑎𝑛𝑑 𝑙𝑖𝑚𝑖𝑡 𝑖𝑠 − ∞
𝑎𝑛𝑑 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑 𝑙𝑖𝑚𝑖𝑡 𝑖𝑠 + ∞)

45. Methods to find Limit
If the limit of a function is in indeterminate form, we can use following
ways to find the limits
1.Factorize the given function and solve
2. Multiply the Nr and Dr by the conjugate of (either Nr or Dr) and
solve
3. Use L’Hospital rule. lim
𝑥→𝑎
𝑓 𝑥
𝑔 𝑥
= lim
𝑥→𝑎
𝑓′ 𝑥
𝑔′ 𝑥

46. Problem 1 (Factorization)
Find the limit of 𝐥𝐢𝐦
𝐱→𝟏
𝐱 𝟑−𝟏
𝐱 𝟐−𝟏
Solution:
lim
x→1
x3−1
x2−1
=
13−1
12−1
=
1−1
1−1
=
0
0
(indeterminate)
lim
x→1
x3−1
x2−1
= lim
x→1
(x−1)(x2+1+x)
x2−1
= lim
x→1
(x−1)(x2+1+x)
(x−1)(x+1)
= lim
x→1
(x2+1+x)
x+1
=
1+1+1
1+1
=
3
2
Problem 2 (multiply by Conjugate)
Solve 𝒍𝒊𝒎
𝒙→𝟎
𝟏+𝒙 𝟐− 𝟏−𝒙 𝟐
𝒙
Solution:
lim
𝑥→0
1+𝑥2− 1−𝑥2
𝑥
=
1−1
0
=
0
0
(indeterminate)
lim
𝑥→0
1+𝑥2− 1−𝑥2
𝑥
= lim
𝑥→0
1+𝑥2− 1−𝑥2
𝑥
1+𝑥2+ 1−𝑥2
1+𝑥2+ 1−𝑥2
= lim
𝑥→0
1+𝑥2−(1−𝑥2)
𝑥( 1+𝑥2+ 1−𝑥2)
= lim
𝑥→0
2𝑥2
𝑥( 1+𝑥2+ 1−𝑥2)
= lim
𝑥→0
2𝑥
( 1+𝑥2+ 1−𝑥2)
=
2 0
1+1
= 0
Problem 3
Solve 𝒍𝒊𝒎
𝒙→𝒂
𝒙+𝟐
𝟓
𝟐− 𝒂+𝟐
𝟓
𝟐
𝒙−𝒂
Solution:
lim
𝑥→𝑎
𝑥+2
5
2− 𝑎+2
5
2
𝑥−𝑎
=
𝑎+2
5
2− 𝑎+2
5
2
𝑎−𝑎
=
0
0
(indeterminate)
lim
𝑥→𝑎
𝑥+2
5
2− 𝑎+2
5
2
𝑥−𝑎
= lim
𝑥→𝑎
𝑥+2
5
2− 𝑎+2
5
2
𝑥+2−2−𝑎
= lim
𝑥→𝑎
𝑥+2
5
2− 𝑎+2
5
2
(𝑥+2)−(𝑎+2)
= lim
𝑥→𝑎
𝑥+2
5
2− 𝑎+2
5
2
(𝑥+2)−(𝑎+2)
=
5
2
𝑎 + 2
5
2
−1
(By using lim
𝑥→𝑎
𝑥 𝑛−𝑎 𝑛
𝑥−𝑎
= 𝑛𝑎 𝑛−1
)
=
5
2
𝑎 + 2
3
2

47. Problem 4
Find the value of 𝐥𝐢𝐦
𝐱→𝟎
𝐚 𝐱−𝐛 𝐱
𝐱
Solution:
lim
𝑥→0
𝑎 𝑥
− 𝑏 𝑥
𝑥
=
𝑎0
− 𝑏0
0
=
1 − 1
0
=
0
0
(Indeterminate)
lim
𝑥→0
𝑎 𝑥−𝑏 𝑥
𝑥
= lim
𝑥→0
𝑎 𝑥−1+1−𝑏 𝑥
𝑥
= lim
𝑥→0
𝑎 𝑥−1−(𝑏 𝑥−1)
𝑥
= lim
𝑥→0
(𝑎 𝑥−1)
𝑥
− lim
𝑥→0
(𝑏 𝑥−1)
𝑥
= log 𝑒 𝑎 − log 𝑒 𝑏
(By using lim
𝑥→0
𝑎 𝑥−1
𝑥
= log 𝑒 𝑎 , 𝑎 > 0)
= log 𝑒
𝑎
𝑏
Problem 5 (Using L’ Hospital
Rule)
Solve 𝐥𝐢𝐦
𝒙→𝟎
𝒂 𝒙−𝟏
𝒙
Solution:
lim
𝑥→0
𝑎 𝑥−1
𝑥
=
𝑎0−1
0
=
1−1
0
=
0
0
(indeterminante)
lim
𝑥→0
𝑎 𝑥−1
𝑥
= lim
𝑥→0
𝑎 𝑥 log 𝑎
1
(∵ d ax =
ax
loga & by using LH rule)
= 𝑎0 log 𝑎 = log 𝑎
Problem 6 (Using L’ Hospital
Rule)
Solve 𝐥𝐢𝐦
𝒙→𝟎
𝒔𝒊𝒏𝒙
𝒙
Solution:
lim
𝑥→0
𝑠𝑖𝑛𝑥
𝑥
=
𝑠𝑖𝑛0
0
=
0
0
(indeterminante)
lim
𝑥→0
𝑠𝑖𝑛𝑥
𝑥
= lim
𝑥→0
𝑐𝑜𝑠𝑥
1
(∵ d 𝑠𝑖𝑛𝑥 =
𝑐𝑜𝑠𝑥 & by using LH rule)
=
𝑐𝑜𝑠0
1
=
1
1
= 1

48. Limits at Infinity
If the function is in indeterminate
form when the limit approaches ∞,
then use the following ideas,
1. for all n > 0, lim
𝑥→∞
1
𝑥 𝑛 = 0 and
solve
2. L’Hospital Rule
Problem 1
Solve 𝒍𝒊𝒎
𝒙→∞
𝒙 𝟐+𝟐𝒙−𝟒
𝟏𝟐𝒙+𝟑𝟏
Solution:
lim
𝑥→∞
𝑥2+2𝑥−4
12𝑥+31
=
∞
∞
(indeterminate)
lim
𝑥→∞
𝑥2+2𝑥−4
12𝑥+31
= lim
𝑥→∞
𝑥 𝑥+2−
4
𝑥
𝑥(12+
31
𝑥
= lim
𝑥→∞
𝑥+2−
4
𝑥
(12+
31
𝑥
)
=
∞+2−0
12+0
= ∞
Problem 2
Solve 𝒍𝒊𝒎
𝒙→∞
( 𝒙 𝟐 + 𝟏 − 𝒙)
Solution:
lim
𝑥→∞
( 𝑥2 + 1 − 𝑥) = ∞ − ∞ (indeterminate)
lim
𝑥→∞
( 𝑥2 + 1 − 𝑥)
= lim
𝑥→∞
( 𝑥2 + 1 − 𝑥)
( 𝑥2+1+𝑥)
( 𝑥2+1+𝑥)
(Use conjugate)
= lim
𝑥→∞
𝑥2+1−𝑥2
( 𝑥2+1+𝑥)
= lim
𝑥→∞
1
( 𝑥2+1+𝑥)
= lim
𝑥→∞
1
(𝑥 1+
1
𝑥2+𝑥)
= lim
𝑥→∞
1
𝑥( 1+
1
𝑥2+1)
=
1
∞
= 0

49. Problem3 (Using L’ Hospital Rule)
Solve 𝒍𝒊𝒎
𝒙→∞
𝒆 𝒙
𝒙 𝟐
Solution:
lim
𝑥→∞
𝑒 𝑥
𝑥2 =
∞
∞
(indeterminate)
Using L’ Hospital Rule
lim
𝑥→∞
𝑒 𝑥
𝑥2
= lim
𝑥→∞
𝑒 𝑥
2𝑥
= lim
𝑥→∞
𝑒 𝑥
2
=
𝑒∞
2
= ∞

50. Squeeze (Or Sandwich) Theorem
The squeeze theorem states that
if 𝑓(𝑥) ≤ 𝑔(𝑥) ≤ ℎ(𝑥) for all numbers 𝑥,
and lim
𝑥→𝑎
𝑓 𝑥 = lim
𝑥→𝑎
ℎ 𝑥 = 𝐿,
𝑡ℎ𝑒𝑛 lim
𝑥→𝑎
𝑔 𝑥 = 𝐿
Problem 1
Find the limit of 𝐥𝐢𝐦
𝒙→𝟎
𝒙 𝟐 𝐜𝐨𝐬
𝟏
𝒙 𝟐
Solution:
W.k.t −1 ≤ 𝐶𝑜𝑠𝜃 ≤ 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝜃
In particular −1 ≤ 𝐶𝑜𝑠
1
𝑥2 ≤ 1
Multiply the inequality by 𝑥2
−𝑥2
≤ 𝑥2
𝐶𝑜𝑠
1
𝑥2 ≤ 𝑥2
The given function bounded by−𝑥2 𝑎𝑛𝑑 𝑥2
Also lim
𝑥→0
−𝑥2 = 0 & lim
𝑥→0
𝑥2 = 0
Here 𝑓 𝑥 = −𝑥2 , ℎ 𝑥 = 𝑥2 & 𝑔 𝑥 = 𝑥2 cos
1
𝑥2
By Squeeze Theorem
lim
𝑥→0
𝑥2 cos
1
𝑥2 = 0

51. Continuous Function
A function 𝑓(𝑥) in a single variable 𝑥 is said to be continuous at point ‘𝑎’ if
1. 𝑓(𝑎) is defined, so that a is in the domain of 𝑓.
2. 𝑙𝑖𝑚 𝑥→ 𝑎 𝑓(𝑥) , exists for 𝑥 in the domain of 𝑓.
3. 𝑙𝑖𝑚 𝑥→𝑎 𝑓 𝑥 = 𝑓 𝑎
In other words, a function 𝑓 is continuous at a point 𝑥 = 𝑎, when
1. The function 𝑓 is defined at 𝑎,
2. The limit of f as 𝑥 approaches a from the right-hand and left-hand limits exist
and are equal,
3. The limit of 𝑓 as 𝑥 approaches a is equal to 𝑓(𝑎)

52. A continuous function is a real-valued function whose graph does not have any
breaks or holes.
Here f(x) is continuous without any break whereas g(x) is not

53. A function which is not continuous is called discontinuous function.
In other words, a function 𝑓 is discontinuous at a point 𝑥 = 𝑎, if it is not satisfying
any one or more conditions (condition 1,2,3 for a continuous function)
Problem1
Check whether the function 𝒇 𝒙 =
𝒙 + 𝟐 𝒊𝒇 𝒙 > 𝟏
𝟏 𝒊𝒇 𝒙 ≤ 𝟏
is continuous or not at 𝒙 = 𝟏.
Solution:
R.H.L
lim
𝑥→1+
𝑓 𝑥 = lim
𝑥→1+
𝑥 + 2 = 1 + 2 = 3
L.H.L
lim
𝑥→1−
𝑓 𝑥 = lim
𝑥→1−
1 = 1
⟹ lim
𝑥→1+
𝑓 𝑥 ≠ lim
𝑥→1−
𝑓 𝑥
Hence it is not continuous at 𝑥 = 1, but it is continuous everywhere else

54. Problem 2
Check whether the function is continuous at x = 0 or not
𝐟 𝐱 =
−𝟏 𝐢𝐟 𝐱 ≤ 𝟎
𝟏 𝐢𝐟 𝐱 > 𝟎
Solution:
R.H.L
lim
x→0+
f x = lim
x→0+
1 = 1
L.H.L
lim
x→0−
f x = lim
x→0−
−1 = −1
⟹ lim
𝑥→0+
𝑓 𝑥 ≠ lim
𝑥→0−
𝑓 𝑥
Hence it is not continuous at 𝑥 = 0

55. Problem 3
Show that the function
𝒇 𝒙 =
−𝒙, 𝒘𝒉𝒆𝒏 𝒙 ≤ 𝟎
𝒙, 𝒘𝒉𝒆𝒏 𝟎 < 𝒙 < 𝟏
𝟐 − 𝒙, 𝒘𝒉𝒆𝒏 𝒙 ≥ 𝟏
is continuous at x=0 and x=1
Solution:
R.H.L
lim
x→0+
f x = lim
x→0+
x = 0
L.H.L
lim
x→0−
f x = lim
x→0−
−x = 0
⟹ lim
𝑥→0+
𝑓 𝑥 = lim
𝑥→0−
𝑓 𝑥
Hence it is continuous at 𝑥 = 0
R.H.L
lim
𝑥→1+
𝑓 𝑥 = lim
𝑥→1+
2 − 𝑥 = 2 − 1 = 1
L.H.L
lim
𝑥→1−
𝑓 𝑥 = lim
𝑥→1−
𝑥 = 1
⟹ lim
𝑥→1+
𝑓 𝑥 = lim
𝑥→1−
𝑓 𝑥
Hence it is continuous at 𝑥 = 1

56. Problem 4
Show that 𝒇 𝒙 =
𝒆
𝟏
𝒙−𝟏
𝒆
𝟏
𝒙+𝟏
when 𝒙 = 𝟎,
𝒇(𝟎) = 𝟏 is discontinuous at 𝒙 = 𝟎
Solution:
RHL
As 𝑥 → 0+
,
1
𝑥
→ +∞
W.k.t, 𝑒∞
= ∞ & 𝑒−∞
= 0
∴ 𝑤ℎ𝑒𝑛 𝑥 ⟶ 0+
,
𝑒
1
𝑥 = 𝑒∞
= ∞ & 𝑒−
1
𝑥 = 𝑒−∞
= 0
lim
𝑥→0+
𝑓 𝑥 = lim
𝑥→0+
𝑒
1
𝑥−1
𝑒
1
𝑥+1
Divide the Nr & Dr by 𝑒
1
𝑥
= lim
𝑥→0+
𝑒
1
𝑥
𝑒
1
𝑥
−
1
𝑒
1
𝑥
𝑒
1
𝑥
𝑒
1
𝑥
+
1
𝑒
1
𝑥
= lim
𝑥→0+
1−
1
𝑒
1
𝑥
1+
1
𝑒
1
𝑥
= lim
𝑥→0+
1−𝑒
−
1
𝑥
1+𝑒
−
1
𝑥
=
1−𝑒−∞
1+𝑒−∞
=
1−0
1+0
=
1
1
= 1
∴ lim
𝑥→0+
𝑓 𝑥 = 1
LHL
As 𝑥 → 0−
,
1
𝑥
→ −∞
W.k.t, 𝑒∞
= ∞ & 𝑒−∞
= 0
∴ 𝑤ℎ𝑒𝑛 𝑥 ⟶ 0−
,
𝑒
1
𝑥 = 𝑒−∞
= 0 & 𝑒−
1
𝑥 = 𝑒∞
= ∞
lim
𝑥→0−
𝑓 𝑥 = lim
𝑥→0−
𝑒
1
𝑥−1
𝑒
1
𝑥+1
= lim
𝑥→0−
𝑒
1
𝑥−1
𝑒
1
𝑥+1
=
𝑒−∞−1
𝑒−∞+1
=
0−1
0+1
= −
1
1
= −1
∴ lim
𝑥→0−
𝑓 𝑥 = −1
⟹ lim
𝑥→0+
𝑓 𝑥 ≠ lim
𝑥→0−
𝑓 𝑥
Hence it is not continuous at 𝑥 = 0