L3 Limit and Limit Theorems

Limit of a Function and Limit Theorems
Mathematics 100
Institute of Mathematics
Math 100 (Institute of Mathematics) Limit and Limit Theorems 1 / 23

Outline
1 Limit of a Function
2 Limit Theorems
3 Lecture Exercises

Limit of a Function
2x2 + x − 3
Consider f (x) = .
x−1

Limit of a Function
2x2 + x − 3
Consider f (x) = .
x−1
What can we say about values of f (x) for values of x near 1 but not equal to
1?

Limit of a Function
2x2 + x − 3
Consider f (x) = .
x−1
What can we say about values of f (x) for values of x near 1 but not equal to
1?
Note that
2x2 + x − 3
f (x) =
x−1
(2x + 3)(x − 1)
=
x−1
= 2x + 3, x 6= 1

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x)

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x)
0

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x)
0 3

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x)
0 3
0.25

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x)
0 3
0.25 3.5

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x)
0 3
0.25 3.5
0.5 4
0.75 4.5
0.9 4.8
0.99 4.98
0.999 4.998
0.9999 4.9998
0.99999 4.99998

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x) x f (x)
0 3
0.25 3.5
0.5 4
0.75 4.5
0.9 4.8
0.99 4.98
0.999 4.998
0.9999 4.9998
0.99999 4.99998

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x) x f (x)
0 3 2
0.25 3.5
0.5 4
0.75 4.5
0.9 4.8
0.99 4.98
0.999 4.998
0.9999 4.9998
0.99999 4.99998

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x) x f (x)
0 3 2 7
0.25 3.5
0.5 4
0.75 4.5
0.9 4.8
0.99 4.98
0.999 4.998
0.9999 4.9998
0.99999 4.99998

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x) x f (x)
0 3 2 7
0.25 3.5 1.75
0.5 4
0.75 4.5
0.9 4.8
0.99 4.98
0.999 4.998
0.9999 4.9998
0.99999 4.99998

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x) x f (x)
0 3 2 7
0.25 3.5 1.75 6.5
0.5 4
0.75 4.5
0.9 4.8
0.99 4.98
0.999 4.998
0.9999 4.9998
0.99999 4.99998

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x) x f (x)
0 3 2 7
0.25 3.5 1.75 6.5
0.5 4 1.5 6
0.75 4.5 1.25 5.5
0.9 4.8 1.1 5.2
0.99 4.98 1.01 5.02
0.999 4.998 1.001 5.002
0.9999 4.9998 1.0001 5.0002
0.99999 4.99998 1.00001 5.00002

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x) x f (x)
0 3 2 7
0.25 3.5 1.75 6.5
0.5 4 1.5 6
0.75 4.5 1.25 5.5
0.9 4.8 1.1 5.2
0.99 4.98 1.01 5.02
0.999 4.998 1.001 5.002
0.9999 4.9998 1.0001 5.0002
0.99999 4.99998 1.00001 5.00002
Based on the tables, even though f (x) is undefined at x = 1,

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x) x f (x)
0 3 2 7
0.25 3.5 1.75 6.5
0.5 4 1.5 6
0.75 4.5 1.25 5.5
0.9 4.8 1.1 5.2
0.99 4.98 1.01 5.02
0.999 4.998 1.001 5.002
0.9999 4.9998 1.0001 5.0002
0.99999 4.99998 1.00001 5.00002
Based on the tables, even though f (x) is undefined at x = 1, the values of

f (x) get closer and closer to 5

Limit of a Function
2x2 + x − 3
f (x) = = 2x + 3, x 6= 1
x−1
x f (x) x f (x)
0 3 2 7
0.25 3.5 1.75 6.5
0.5 4 1.5 6
0.75 4.5 1.25 5.5
0.9 4.8 1.1 5.2
0.99 4.98 1.01 5.02
0.999 4.998 1.001 5.002
0.9999 4.9998 1.0001 5.0002
0.99999 4.99998 1.00001 5.00002
Based on the tables, even though f (x) is undefined at x = 1, the values of

f (x) get closer and closer to 5 as we take values of x very near 1.

Limit of a Function
Definition
The limit of a function f (x) as x approaches a is L
if the values of f (x) get closer and closer to L as x assumes values getting
closer and closer to a but not reaching a.

Limit of a Function
Definition
We write
lim f (x) = L.
x→a

Limit of a Function
Definition
We write
lim f (x) = L.
x→a
Remark:
Note that if the limit of a function exists, then it is unique.

Limit of a Function
The limit of the function may also be observed from its graph.
−2 −1 1 2 3 4
−1
−2
2x2 + x − 3
Graph of f (x) =
x−1

Limit of a Function
lim f (x) = 5
x→1
6
−2 −1 1 2 3 4
−1
−2
2x2 + x − 3
Graph of f (x) =
x−1

Limit of a Function
lim f (x) = 5
x→1
6
5
Note: lim f (x) exists even
4 x→1
though f (1) is undefined.
3
−2 −1 1 2 3 4
−1
−2
2x2 + x − 3
Graph of f (x) =
x−1

Properties of Limits
Theorem
Let a, b ∈ R and n a positive integer. Then

Theorem
Let a, b ∈ R and n a positive integer. Then
(i) lim b
x→a

Theorem
R
Let a, b ∈ and n a positive integer. Then
(i) lim b = b
x→a

Theorem
R
(i) lim b = b
x→a
(ii) lim x
x→a

Theorem
R
(i) lim b = b
x→a
(ii) lim x = a
x→a

Theorem
R
(i) lim b = b
x→a
(ii) lim x = a
x→a
(iii) lim xn = an
x→a

Theorem
R
(i) lim b = b
x→a
(ii) lim x = a
x→a
(iii) lim xn = an
x→a
√ √
(iv) lim n x = n a, provided that a > 0 when n is even
x→a

Operations with Limits
Theorem
Let a, b ∈ R and n be a positive integer. If x→a
lim f (x) and lim g(x) both exist,
x→a
then the following operations are valid:

Theorem
x→a
(i) lim (f (x) ± g(x))
x→a

Theorem
x→a
(i) lim (f (x) ± g(x)) = lim f (x) ± lim g(x)
x→a x→a x→a

Theorem
x→a
x→a x→a x→a
(ii) lim (f (x) · g(x))
x→a

Theorem
x→a
x→a x→a x→a
(ii) lim (f (x) · g(x)) = lim f (x) · lim g(x)
x→a x→a x→a

Theorem
x→a
x→a x→a x→a
x→a x→a x→a
(iii) lim (bf (x))
x→a

Theorem
x→a
x→a x→a x→a
x→a x→a x→a
(iii) lim (bf (x)) = b · lim f (x)
x→a x→a

Theorem
x→a
x→a x→a x→a
x→a x→a x→a
x→a x→a
f (x)
(iv) If lim g(x) 6= 0, then lim
x→a x→a g(x)

Theorem
x→a
x→a x→a x→a
x→a x→a x→a
x→a x→a
f (x) lim f (x)

(iv) If lim g(x) 6= 0, then lim = x→a
x→a x→a g(x) lim g(x)
x→a

Theorem
x→a
x→a x→a x→a
x→a x→a x→a
x→a x→a
f (x) lim f (x)

x→a
n
(v) lim (f (x))
x→a

Theorem
x→a
x→a x→a x→a
x→a x→a x→a
x→a x→a
f (x) lim f (x)

x→a
n
n
(v) lim (f (x)) = lim f (x)
x→a x→a

Theorem
x→a
x→a x→a x→a
x→a x→a x→a
x→a x→a
f (x) lim f (x)

x→a
n
n
x→a x→a
p
(vi) If n is even and lim f (x) > 0, then lim n f (x)
x→a x→a

Theorem
x→a
x→a x→a x→a
x→a x→a x→a
x→a x→a
f (x) lim f (x)

x→a
n
n
x→a x→a
p q
(vi) If n is even and lim f (x) > 0, then lim n f (x) = n lim f (x)
x→a x→a x→a

The limit of a polynomial function
Theorem
If P (x) is a polynomial function and a ∈ R, then x→a
lim P (x) = P (a).

The Replacement Theorem
Theorem
Let a ∈ R such that f (x) = g(x) for all x 6= a. If x→a
lim g(x) exists, then lim f (x)
x→a
also exists, and
lim f (x) = lim g(x).
x→a x→a

Evaluating Limits
Example
Evaluate lim 3.
x→2
Solution:

Evaluating Limits
Example
Evaluate lim 3.
x→2
Solution:
We simply apply the property on limit of a constant:
lim 3
x→2

Evaluating Limits
Example
Evaluate lim 3.
x→2
Solution:
lim 3 = 3.
x→2

Evaluating Limits
Example
Evaluate lim 3.
x→2
Solution:
lim 3 = 3.
x→2
Example
Evaluate lim (2x2 − 5x + 3).
x→2
Solution:

Evaluating Limits
Example
Evaluate lim 3.
x→2
Solution:
lim 3 = 3.
x→2
Example
x→2
Solution:
Apply the theorem on the limit of a polynomial:
lim (2x2 − 5x + 3) =
x→2

Evaluating Limits
Example
Evaluate lim 3.
x→2
Solution:
lim 3 = 3.
x→2
Example
x→2
Solution:
lim (2x2 − 5x + 3) = 2(2)2 −

x→2

Evaluating Limits
Example
Evaluate lim 3.
x→2
Solution:
lim 3 = 3.
x→2
Example
x→2
Solution:
lim (2x2 − 5x + 3) = 2(2)2 − 5(2) +

x→2

Evaluating Limits
Example
Evaluate lim 3.
x→2
Solution:
lim 3 = 3.
x→2
Example
x→2
Solution:
lim (2x2 − 5x + 3) = 2(2)2 − 5(2) + 3

x→2

Evaluating Limits
Example
Evaluate lim 3.
x→2
Solution:
lim 3 = 3.
x→2
Example
x→2
Solution:
lim (2x2 − 5x + 3) = 2(2)2 − 5(2) + 3 = 1.

x→2

Evaluating Limits
Example
x2 + 2x
Evaluate lim .
x→3 2x + 3
Solution:

Evaluating Limits
Example
x2 + 2x
Evaluate lim .
x→3 2x + 3
Solution:
By the property on the limit of a quotient:
x2 + 2x
lim
x→3 2x + 3

Evaluating Limits
Example
x2 + 2x
Evaluate lim .
x→3 2x + 3
Solution:
x2 + 2x lim (x2 + 2x)

lim = x→3
x→3 2x + 3

Evaluating Limits
Example
x2 + 2x
Evaluate lim .
x→3 2x + 3
Solution:
x2 + 2x lim (x2 + 2x)

lim = x→3
x→3 2x + 3 lim (2x + 3)
x→3

Evaluating Limits
Example
x2 + 2x
Evaluate lim .
x→3 2x + 3
Solution:
x2 + 2x lim (x2 + 2x) (3)2 + 2(3)

lim = x→3 =
x→3 2x + 3 lim (2x + 3)
x→3

Evaluating Limits
Example
x2 + 2x
Evaluate lim .
x→3 2x + 3
Solution:
x2 + 2x lim (x2 + 2x) (3)2 + 2(3)

lim = x→3 =
x→3 2x + 3 lim (2x + 3) 2(3) + 3
x→3

Evaluating Limits
Example
x2 + 2x
Evaluate lim .
x→3 2x + 3
Solution:
x2 + 2x lim (x2 + 2x) (3)2 + 2(3) 15

lim = x→3 = = =
x→3 2x + 3 lim (2x + 3) 2(3) + 3 9
x→3

Evaluating Limits
Example
x2 + 2x
Evaluate lim .
x→3 2x + 3
Solution:
x2 + 2x lim (x2 + 2x) (3)2 + 2(3) 15 5

lim = x→3 = = = .
x→3 2x + 3 lim (2x + 3) 2(3) + 3 9 3
x→3

Evaluating Limits
Example
s
3 2x2 + x + 5
Evaluate lim .
x→1 3x2 + 19x + 5
Solution:

Evaluating Limits
Example
s
3 2x2 + x + 5
Evaluate lim .
x→1 3x2 + 19x + 5
Solution:
By the properties on the limit of a radical and limit of a quotient:
s
3 2x2 + x + 5
lim
x→1 3x2 + 19x + 5

Evaluating Limits
Example
s
3 2x2 + x + 5
Evaluate lim .
x→1 3x2 + 19x + 5
Solution:
s s
3 2x2 + x + 5 3 2x2 + x + 5
lim = lim
x→1 3x2 + 19x + 5 x→1 3x2 + 19x + 5

Evaluating Limits
Example
s
3 2x2 + x + 5
Evaluate lim .
x→1 3x2 + 19x + 5
Solution:
s s v
u
2 2
2x + x + 5 2x + x + 5
u
3 3 3
lim = lim =
u
3x2 + 19x + 5 x→1 3x2 + 19x + 5
t
x→1

Evaluating Limits
Example
s
3 2x2 + x + 5
Evaluate lim .
x→1 3x2 + 19x + 5
Solution:
v
u lim (2x2 + x + 5)
s s u
2 2
3 2x + x + 5 3 2x + x + 5 3 x→1
lim = lim 2 =t
u
x→1 3x2 + 19x + 5 x→1 3x + 19x + 5

Evaluating Limits
Example
s
3 2x2 + x + 5
Evaluate lim .
x→1 3x2 + 19x + 5
Solution:
v
u lim (2x2 + x + 5)
s s u
2 2
3 2x + x + 5 3 2x + x + 5 3 x→1
lim = lim 2 =t
u
x→1 3x2 + 19x + 5 x→1 3x + 19x + 5 lim (3x2 + 19x + 5)
x→1

Evaluating Limits
Example
s
3 2x2 + x + 5
Evaluate lim .
x→1 3x2 + 19x + 5
Solution:
v
u lim (2x2 + x + 5)
s s u
2 2
3 2x + x + 5 3 2x + x + 5 3 x→1
lim = lim 2 =t
u
x→1 3x2 + 19x + 5 x→1 3x + 19x + 5 lim (3x2 + 19x + 5)
x→1
s
3
=

Evaluating Limits
Example
s
3 2x2 + x + 5
Evaluate lim .
x→1 3x2 + 19x + 5
Solution:
v
u lim (2x2 + x + 5)
s s u
2 2
3 2x + x + 5 3 2x + x + 5 3 x→1
lim = lim 2 =t
u
x→1 3x2 + 19x + 5 x→1 3x + 19x + 5 lim (3x2 + 19x + 5)
x→1
s
2(1)2 + 1 + 5
= 3

Evaluating Limits
Example
s
3 2x2 + x + 5
Evaluate lim .
x→1 3x2 + 19x + 5
Solution:
v
u lim (2x2 + x + 5)
s s u
2 2
3 2x + x + 5 3 2x + x + 5 3 x→1
lim = lim 2 =t
u
x→1 3x2 + 19x + 5 x→1 3x + 19x + 5 lim (3x2 + 19x + 5)
x→1
s
2(1)2 + 1 + 5
= 3
3(1)2 + 19(1) + 5

Evaluating Limits
Example
s
3 2x2 + x + 5
Evaluate lim .
x→1 3x2 + 19x + 5
Solution:
v
u lim (2x2 + x + 5)
s s u
2 2
3 2x + x + 5 3 2x + x + 5 3 x→1
lim = lim 2 =t
u
x→1 3x2 + 19x + 5 x→1 3x + 19x + 5 lim (3x2 + 19x + 5)
x→1
s r
3 2(1)2 + 1 + 5 3 8
= =
3(1)2 + 19(1) + 5 27

Evaluating Limits
Example
s
3 2x2 + x + 5
Evaluate lim .
x→1 3x2 + 19x + 5
Solution:
v
u lim (2x2 + x + 5)
s s u
2 2
3 2x + x + 5 3 2x + x + 5 3 x→1
lim = lim 2 =t
u
x→1 3x2 + 19x + 5 x→1 3x + 19x + 5 lim (3x2 + 19x + 5)
x→1
s r
3 2(1)2 + 1 + 5 3 8 2
= = = .
3(1)2 + 19(1) + 5 27 3

Evaluating Limits
Example
2x2 − 5x − 3
Evaluate lim .
x→3 3−x
Solution:

Evaluating Limits
Example
2x2 − 5x − 3
Evaluate lim .
x→3 3−x
Solution:
We first note that if we substitute x = 3 in the given expression directly, we
would get

Evaluating Limits
Example
2x2 − 5x − 3
Evaluate lim .
x→3 3−x
Solution:
would get
2(3)2 −

Evaluating Limits
Example
2x2 − 5x − 3
Evaluate lim .
x→3 3−x
Solution:
would get
2(3)2 − 5(3) −

Evaluating Limits
Example
2x2 − 5x − 3
Evaluate lim .
x→3 3−x
Solution:
would get
2(3)2 − 5(3) − 3

Evaluating Limits
Example
2x2 − 5x − 3
Evaluate lim .
x→3 3−x
Solution:
would get
2(3)2 − 5(3) − 3
3 − (3)

Evaluating Limits
Example
2x2 − 5x − 3
Evaluate lim .
x→3 3−x
Solution:
would get
2(3)2 − 5(3) − 3 18 − 15 − 3
=
3 − (3) 0

Evaluating Limits
Example
2x2 − 5x − 3
Evaluate lim .
x→3 3−x
Solution:
would get
2(3)2 − 5(3) − 3 18 − 15 − 3 0
= = .
3 − (3) 0 0

Evaluating Limits
Example
2x2 − 5x − 3
Evaluate lim .
x→3 3−x
Solution:
would get
2(3)2 − 5(3) − 3 18 − 15 − 3 0
= = .
3 − (3) 0 0
0
The expression is not a number! It is called an indeterminate form.
0

Evaluating Limits
Example
2x2 − 5x − 3
Evaluate lim .
x→3 3−x
Solution:
would get
2(3)2 − 5(3) − 3 18 − 15 − 3 0
= = .
3 − (3) 0 0
0
The expression is not a number! It is called an indeterminate form.
0
Thus, direct substitution cannot be applied to evaluate this limit.

Evaluating Limits
Solution cont’d.:

Evaluating Limits
Solution cont’d.:
We observe that when x 6= 3,
2x2 − 5x − 3
3−x

Evaluating Limits
Solution cont’d.:
2x2 − 5x − 3 (2x + 1)(x − 3)

=
3−x −(x − 3)

Evaluating Limits
Solution cont’d.:
2x2 − 5x − 3 (2x + 1)(x − 3)

= = −(2x + 1) = −2x − 1.
3−x −(x − 3)

Evaluating Limits
Solution cont’d.:
2x2 − 5x − 3 (2x + 1)(x − 3)

= = −(2x + 1) = −2x − 1.
3−x −(x − 3)
Thus, by the replacement theorem,
2x2 − 5x − 3
lim
x→3 3−x

Evaluating Limits
Solution cont’d.:
2x2 − 5x − 3 (2x + 1)(x − 3)

= = −(2x + 1) = −2x − 1.
3−x −(x − 3)
2x2 − 5x − 3
lim = lim (−2x − 1)
x→3 3−x x→3

Evaluating Limits
Solution cont’d.:
2x2 − 5x − 3 (2x + 1)(x − 3)

= = −(2x + 1) = −2x − 1.
3−x −(x − 3)
2x2 − 5x − 3
lim = lim (−2x − 1) = −2(3) − 1
x→3 3−x x→3

Evaluating Limits
Solution cont’d.:
2x2 − 5x − 3 (2x + 1)(x − 3)

= = −(2x + 1) = −2x − 1.
3−x −(x − 3)
2x2 − 5x − 3
lim = lim (−2x − 1) = −2(3) − 1 = −7.
x→3 3−x x→3

Evaluating Limits
Example
√ √
2+x− 2
Evaluate lim .
x→0 x

Evaluating Limits
Example
√ √
2+x− 2 0
Evaluate lim .
x→0 x 0
Solution:

Evaluating Limits
Example
√ √
2+x− 2 0
Evaluate lim .
x→0 x 0
Solution:
Note that if x 6= 0, then
√ √
2+x− 2
x

Evaluating Limits
Example
√ √
2+x− 2 0
Evaluate lim .
x→0 x 0
Solution:
√ √ √ √ √ √
2+x− 2 2+x− 2 2+x+ 2
= ·√ √
x x 2+x+ 2

Evaluating Limits
Example
√ √
2+x− 2 0
Evaluate lim .
x→0 x 0
Solution:
√ √ √ √ √ √
2+x− 2 2+x− 2 2+x+ 2
= ·√ √
x x 2+x+ 2
(2 + x) − 2
= √ √
x( 2 + x + 2)

Evaluating Limits
Example
√ √
2+x− 2 0
Evaluate lim .
x→0 x 0
Solution:
√ √ √ √ √ √
2+x− 2 2+x− 2 2+x+ 2
= ·√ √
x x 2+x+ 2
(2 + x) − 2
= √ √
x( 2 + x + 2)
x
= √ √
x( 2 + x + 2)

Evaluating Limits
Example
√ √
2+x− 2 0
Evaluate lim .
x→0 x 0
Solution:
√ √ √ √ √ √
2+x− 2 2+x− 2 2+x+ 2
= ·√ √
x x 2+x+ 2
(2 + x) − 2
= √ √
x( 2 + x + 2)
x
= √ √
x( 2 + x + 2)
1
=√ √
2+x+ 2

Evaluating Limits
Solution cont’d.:

Evaluating Limits
Solution cont’d.:
Thus, by the replacement theorem we have
√ √
2+x− 2
lim
x→0 x

Evaluating Limits
Solution cont’d.:
√ √
2+x− 2 1
lim = lim √ √
x→0 x x→0 2+x+ 2

Evaluating Limits
Solution cont’d.:
√ √
2+x− 2 1 1
lim = lim √ √ =√ √
x→0 x x→0 2+x+ 2 2+0+ 2

Evaluating Limits
Solution cont’d.:
√ √
2+x− 2 1 1 1
lim = lim √ √ =√ √ = √ .
x→0 x x→0 2+x+ 2 2+0+ 2 2 2

Evaluating Limits
Example
x3 + 8
Evaluate lim .
x→−2 x + 2

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:
Recall that a3 + b3

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:
Recall that a3 + b3 = (a + b)

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:
Recall that a3 + b3 = (a + b)(a2

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:
Recall that a3 + b3 = (a + b)(a2 − ab

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:
Recall that a3 + b3 = (a + b)(a2 − ab + b2 ).

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:
We observe that for x 6= −2,
x3 + 8
x+2

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:
x3 + 8 (x + 2)(x2 − 2x + 4)
=
x+2 x+2

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:
x3 + 8 (x + 2)(x2 − 2x + 4)
= = x2 − 2x + 4.
x+2 x+2

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:
x3 + 8 (x + 2)(x2 − 2x + 4)
= = x2 − 2x + 4.
x+2 x+2
Thus,
x3 + 8
lim = lim (x2 − 2x + 4)
x→−2 x + 2 x→−2

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:
x3 + 8 (x + 2)(x2 − 2x + 4)
= = x2 − 2x + 4.
x+2 x+2
Thus,
x3 + 8
lim = lim (x2 − 2x + 4)
x→−2 x + 2 x→−2
= (−2)2 − 2(−2) + 4

Evaluating Limits
Example
x3 + 8

0
Evaluate lim .
x→−2 x + 2 0
Solution:
x3 + 8 (x + 2)(x2 − 2x + 4)
= = x2 − 2x + 4.
x+2 x+2
Thus,
x3 + 8
lim = lim (x2 − 2x + 4)
x→−2 x + 2 x→−2
= (−2)2 − 2(−2) + 4
= 12.

Evaluating Limits
Example
√
3−x−1
Evaluate lim .
x→2 x−2

Evaluating Limits
Example
√
3−x−1 0
Evaluate lim .
x→2 x−2 0
Solution:

Evaluating Limits
Example
√
3−x−1 0
Evaluate lim .
x→2 x−2 0
Solution:
For x 6= 2,
√ √ √
3−x−1 3−x−1 3−x+1
= ·√
x−2 x−2 3−x+1

Evaluating Limits
Example
√
3−x−1 0
Evaluate lim .
x→2 x−2 0
Solution:
For x 6= 2,
√ √ √
3−x−1 3−x−1 3−x+1
= ·√
x−2 x−2 3−x+1
(3 − x) − 1
= √
(x − 2)( 3 − x + 1)

Evaluating Limits
Example
√
3−x−1 0
Evaluate lim .
x→2 x−2 0
Solution:
For x 6= 2,
√ √ √
3−x−1 3−x−1 3−x+1
= ·√
x−2 x−2 3−x+1
(3 − x) − 1
= √
(x − 2)( 3 − x + 1)
−(x − 2)
= √
(x − 2)( 3 − x + 1)

Evaluating Limits
Example
√
3−x−1 0
Evaluate lim .
x→2 x−2 0
Solution:
For x 6= 2,
√ √ √
3−x−1 3−x−1 3−x+1
= ·√
x−2 x−2 3−x+1
(3 − x) − 1
= √
(x − 2)( 3 − x + 1)
−(x − 2)
= √
(x − 2)( 3 − x + 1)
1
= −√
3−x+1

Evaluating Limits
Solution cont’d.:

Evaluating Limits
Solution cont’d.:
Hence,
√
3−x−1
lim
x→2 x−2

Evaluating Limits
Solution cont’d.:
Hence,
√
3−x−1 1
lim = lim − √
x→2 x−2 x→2 3−x+1

Evaluating Limits
Solution cont’d.:
Hence,
√
3−x−1 1 1
lim = lim − √ = −√
x→2 x−2 x→2 3−x+1 3−2+1

Evaluating Limits
Solution cont’d.:
Hence,
√
3−x−1 1 1 1
lim = lim − √ = −√ =− .
x→2 x−2 x→2 3−x+1 3−2+1 2

Evaluating Limits
Example
√
3
x−1
Evaluate lim .
x→1 x−1

Evaluating Limits
Example
√
3
x−1 0
Evaluate lim .
x→1 x−1 0
Solution:

Evaluating Limits
Example
√
3
x−1 0
Evaluate lim .
x→1 x−1 0
Solution:
If x 6= 1, then
√ √
3
x−1 3
x−1
= ·
x−1 x−1

Evaluating Limits
Example
√
3
x−1 0
Evaluate lim .
x→1 x−1 0
Solution:
If x 6= 1, then
√ √ √
3
3
x−1 3
x−1 x2
= ·
x−1 x−1

Evaluating Limits
Example
√
3
x−1 0
Evaluate lim .
x→1 x−1 0
Solution:
If x 6= 1, then
√ √ √
3 √
3
x−1 3
x−1 x2 + 3 x
= ·
x−1 x−1

Evaluating Limits
Example
√
3
x−1 0
Evaluate lim .
x→1 x−1 0
Solution:
If x 6= 1, then
√ √ √
3 √
3
x−1 3
x−1 x2 + 3 x + 1
= ·
x−1 x−1

Evaluating Limits
Example
√
3
x−1 0
Evaluate lim .
x→1 x−1 0
Solution:
If x 6= 1, then
√ √ √
3 √
3
x−1 3
x−1 x2 + 3 x + 1
= · √ √
x−1 x−1 3
x2 + 3 x + 1

Evaluating Limits
Example
√
3
x−1 0
Evaluate lim .
x→1 x−1 0
Solution:
If x 6= 1, then
√ √ √
3 √
3
x−1 3
x−1 x2 + 3 x + 1
= · √ √
x−1 x−1 3
x2 + 3 x + 1
x−1
=

Evaluating Limits
Example
√
3
x−1 0
Evaluate lim .
x→1 x−1 0
Solution:
If x 6= 1, then
√ √ √
3 √
3
x−1 3
x−1 x2 + 3 x + 1
= · √ √
x−1 x−1 3
x2 + 3 x + 1
x−1
= √3 √
(x − 1)( x2 + 3 x + 1)

Evaluating Limits
Example
√
3
x−1 0
Evaluate lim .
x→1 x−1 0
Solution:
If x 6= 1, then
√ √ √
3 √
3
x−1 3
x−1 x2 + 3 x + 1
= · √ √
x−1 x−1 3
x2 + 3 x + 1
x−1
= √3 √
(x − 1)( x2 + 3 x + 1)
1
= √3 √ .
2
x + 3x+1

Evaluating Limits
Solution cont’d.:

Evaluating Limits
Solution cont’d.:
So by the replacement theorem,
√
3
x−1
lim
x→1 x − 1

Evaluating Limits
Solution cont’d.:
√
3
x−1 1
lim = lim √ √
x→1 x − 1 x→1 3 2
x + 3x+1

Evaluating Limits
Solution cont’d.:
√
3
x−1 1 1
lim = lim √ √ = √ √
x→1 x − 1 x→1 3 2
x + x+1
3 3
1 + 31+1
2

Evaluating Limits
Solution cont’d.:
√
3
x−1 1 1 1
lim = lim √ √ = √ √ = .
x→1 x − 1 x→1 3 2
x + x+1
3 3 2 3
1 + 1+1 3

Lecture Exercises
Evaluate the following limits.

s
x2 + 3x + 11
1. lim
x→−1 x3 − x2 − x + 10
2x2 − 3x − 2
2. lim
x→2 x−2
√3
x+7−2
3. lim
x→1 1−x

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Limit of a Function and Limit Theorems

Math 100 (Institute of Mathematics) Limit and Limit Theorems 1 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 2 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 3 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 3 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 3 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Based on the tables, even though f (x) is undefined at x = 1,

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Based on the tables, even though f (x) is undefined at x = 1, the values of

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Based on the tables, even though f (x) is undefined at x = 1, the values of

Math 100 (Institute of Mathematics) Limit and Limit Theorems 4 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 5 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 5 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 5 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 6 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 6 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 6 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 7 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 7 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 7 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 7 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 7 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 7 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 7 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

f (x) lim f (x)

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

f (x) lim f (x)

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

f (x) lim f (x)

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

f (x) lim f (x)

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

f (x) lim f (x)

Math 100 (Institute of Mathematics) Limit and Limit Theorems 8 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 9 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 10 / 23

Math 100 (Institute of Mathematics) Limit and Limit Theorems 11 / 23