Limit and Continuity
1.1 Definition of Limit
1.2 One sided Limit
1.3 Infinite limit and Vertical Asymptotes
1.4 Calculating Limit using Limit Law
1.5 Limit at Infinity and Horizontal Asymptotes
1.5 Limit of Trigonometric Function
1.6 Continuity
1 LIMIT AND CONTINUITY
1.1 DEFINITION OF LIMIT
The most basic use of limits is to describe how a function behaves as the independent variable
approaches a given value. Here we intend to give numerical and graphical approaches to the
concept of limit using examples
Let 𝑓(𝑥) = 2 𝑥 + 2 and compute 𝑓(𝑥) as 𝑥 takes values closer to 1. We first consider
values of 𝑥 approaching 1 from the left (𝑥 < 1).
Solution
𝑥
𝑓(𝑥)
0.5
3
0.8
3.6
0.9
3.8
0.95
3.9
0.99
3.98
0.999
3.998
0.9999
3.9998
0.99999
3.99998
We now consider 𝑥 approaching 1 from the right (𝑥 > 1).
𝑥
𝑓(𝑥)
1.5
5
1.2
4.4
1.1
4.2
1.05
4.1
1.01
4.02
1.001
4.002
1.0001
4.0002
1.00001
4.00002
In both cases as 𝑥 approaches 1, 𝑓(𝑥) approaches 4. Intuitively, we say that
lim 𝑓(𝑥) = 4
𝑥→1
.
LIMIT (DEFINITION)
We are talking about the values that 𝑓(𝑥) takes when 𝑥 gets closer to 1 and not 𝑓(1).
If the values of 𝑓(𝑥) can be made as close as we like to 𝐿 by taking values of 𝑥
sufficiently close to 𝑎
(but not equal to 𝑎), then we write
lim 𝑓(𝑥) = 𝐿
𝑥→𝑎
Which is read “the limit of 𝑓(𝑥) as x approaches a is 𝐿” or “ 𝑓(𝑥) approaches 𝐿
as 𝑥 approaches 𝑎. the expression 𝑓(𝑥) → 𝐿 𝑎𝑠 𝑥 → 𝑎 .
Example 1
For the functions below, graph and find
lim 𝑓(𝑥)
𝑥→1
𝑎)
𝑏)
𝑐)
Solution
Example 2
𝑓(𝑥) =
𝑥2 − 1
𝑥−1
𝑥2 − 1
𝑔(𝑥) = { 𝑥 − 1 ,
1,
ℎ(𝑥) = 𝑥 + 1
𝑥≠1
𝑥=1
Discuss the behavior of the following function as 𝑥 → 0
𝑎)
𝑏)
0, 𝑥 < 0
𝑓(𝑥) = {
1, 𝑥 ≥ 0
1
ℎ(𝑥) = {𝑥 , 𝑥 ≠ 0
0, 𝑥 = 0
Solution
a)
b)
Example 3
For the function 𝑓(𝑡), find the following limits or explain why they don’t exist.
𝑎)
𝑏)
𝑐)
lim 𝑓(𝑡)
𝑡→−2
lim 𝑓(𝑡)
𝑡→−1
lim 𝑓(𝑡)
𝑡→0
Example 4
Which of the following statements about the function 𝑦 = 𝑓(𝑥) graphed here are true and which are
false?
a) lim 𝑓(𝑥) 𝑒𝑥𝑖𝑠𝑡
𝑥→0
b)
lim 𝑓(𝑥) = 1
𝑥→0
c)
lim 𝑓(𝑥) = 0
𝑥→0
d)
lim 𝑓(𝑥) = 1
𝑥→1
e)
lim 𝑓(𝑥) = 0
𝑥→1
Example 5
Explain why the limit s do not exist
𝑥
𝑥→0 |𝑥|
𝑎) lim
b)
1
𝑥→1 𝑥 − 1
lim
1.2 ONE SIDED LIMIT
For some functions, it is appropriate to look at their behavior from one side only. If x
approaches c from the right only, you write
𝐥𝐢𝐦 𝐟(𝐱)
𝐱→𝐜 +
or if x approaches c from the left only, you write
𝐥𝐢𝐦 𝐟(𝐱)
𝐱→𝐜 −
It follows, then, that
𝐥𝐢𝐦 𝐟(𝐱) = 𝐋
𝐱→𝐜
if and only if
𝐥𝐢𝐦 𝒇(𝒙) = 𝐥𝐢𝐦− 𝒇(𝒙) = 𝑳
𝒙→𝒄+
𝒙→𝒄
Example 1: Evaluate
lim √𝑥
𝑥→0+
Because 𝑥 is approaching 0 from the right, it is always positive;√𝑥 is getting closer and closer to
zero, so
lim+ √𝑥 = 0
𝑥→0
Although substituting 0 for 𝑥 would yield the same answer, the next example illustrates why this
technique is not always appropriate.
Example 2: Evaluate
lim √𝑥
𝑥→0−
Because 𝑥 is approaching 0 from the left, it is always negative, and √𝑥 does not exist. In this
situation,
lim− √𝑥
𝑥→0
DNE. Also, note that the two sided limit
lim √𝑥
𝑥→0
𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 lim− √𝑥 = 0 ≠ lim− √𝑥
Please refer to the function 𝑓(𝑥) = √𝑥 below:
𝑥→0
𝑥→0
Example 3: Evaluate
𝑎) lim−
|𝑥 − 2|
𝑥−2
𝑏) lim+
|𝑥 − 2|
𝑥−2
𝑥→2
𝑥→2
|𝑥 − 2|
𝑥→2 𝑥 − 2
𝑐) lim
Ans:
a) As 𝑥 approaches 2 from the left, 𝑥 − 2 is negative, and | 𝑥 − 2| = − ( 𝑥 − 2); hence,
lim
𝑥→2−
|𝑥 − 2| −(𝑥 − 2)
=
= −1
𝑥−2
𝑥−2
b) As 𝑥 approaches 2 from the right, 𝑥 − 2 is positive, and | 𝑥 − 2| = 𝑥 − 2; hence;
|𝑥 − 2| (𝑥 − 2)
lim+
=
=1
𝑥→2 𝑥 − 2
𝑥−2
c) Because
lim−
𝑥→2
|𝑥 − 2|
|𝑥 − 2|
≠ lim+
𝑥→2 𝑥 − 2
𝑥−2
|𝑥 − 2|
Does Not Exist
𝑥→2 𝑥 − 2
Therefore lim
Example 4
Use the definition of one sided limits to prove the statements below:
𝑎) lim−
𝑥
= −1
|𝑥|
𝑏) lim+
𝑥−2
=1
|𝑥 − 2|
𝑥→0
𝑥→2
EXAMPLE 5
The graph above shows that as 𝑥 approaches 2 from the left, 𝑦 = 𝑓(𝑥) approaches 1 and
this can be written as
lim 𝑓(𝑥) = 1
𝑥→2−
As 𝑥 approaches 2 from the right, 𝑦 = 𝑓(𝑥) approaches 2 and this can be written as
lim∓ 𝑓(𝑥) = 2
𝑥→2
Note that the left and right hand limits and 𝑓(2) = 1.5 are all different.
EXAMPLE 6
This graph shows that
lim 𝑓(𝑥) = 1
𝑥→0−
and
lim 𝑓(𝑥) = 1
𝑥→0+
Note that the left and right hand limits are equal and we can write
lim 𝑓(𝑥) = 1
𝑥→0
In this example, the limit when 𝑥 approaches 0 is equal to 𝑓(0) = 1.
Example 7 : For the function 𝑦 = 𝐹(𝑥) , find
(a)
lim
𝐹(𝑥)
𝑥 → 0−
(b)
lim
𝐹(𝑥)
𝑥 → 0+
(c)
𝐹(0)
(d)
lim
𝐹(𝑥)
𝑥→0
EXAMPLE 8
This graph shows that as 𝑥 approaches − 2 from the left, 𝑓(𝑥) gets smaller and smaller
without bound and there is no limit. We write
lim − 𝑓(𝑥) = −∞
x→−2
As 𝑥 approaches − 2 from the right, 𝑓(𝑥) gets larger and larger without bound and there is
no limit. We write
lim 𝑓(𝑥) = +∞
x→−2+
Note that −∞ and +∞ are symbols and not numbers. These are symbols used to indicate
that the limit does not exist.
1.3 INFINITE LIMITE AND VERTICLE ASYMPTOTES
In this section we will take a look at limits whose value is infinity or minus infinity. These kinds of
limit will show up fairly regularly in later sections and in other courses and so you’ll need to be able
to deal with them when you run across them.
The first thing we should probably do here is to define just what we mean when we say that a limit
has a value of infinity or minus infinity.
Definition
𝐥𝐢𝐦 𝒇(𝒙) = +∞
𝒙→𝒂
if we can make 𝑓(𝑥) arbitrarily large for all 𝑥 sufficiently close to 𝑥 = 𝑎, from
both sides, without actually letting 𝑥 = 𝑎.
𝐥𝐢𝐦 𝒇(𝒙) = −∞
𝒙→𝒂
if we can make 𝑓(𝑥) arbitrarily large and negative for all 𝑥 sufficiently close to
𝑥 = 𝑎, from both sides, without actually letting 𝑥 = 𝑎.
Example 1
1
= +∞
2
𝑥→0 𝑥
a) lim
−1
= −∞
2
𝑥→0 (𝑥 − 1)
b) lim
1
=∞
𝑥→0 𝑥
c) lim
d) lim
𝑥→2+
𝑥−3
𝑥2 − 4
𝐴𝑛𝑠: − ∞
𝑥−3
2
𝑥→2− 𝑥 − 4
𝐴𝑛𝑠: + ∞
𝑥−3
2
𝑥→2 𝑥 − 4
𝐴𝑛𝑠: 𝐷𝑁𝐸
e) lim
f) lim
3 − 𝑥,
1) Let 𝑓(𝑥) = {𝑥
+ 1,
2
𝑥<2
𝑥>2
a) Find lim 𝑓(𝑥) and lim− 𝑓(𝑥)
𝑥→2
𝑥→2+
b) Does lim 𝑓(𝑥)
𝑥→2
exists? If so, what it is? If not, why not?
c) Find lim 𝑓(𝑥)and lim− 𝑓(𝑥)
𝑥→4 +
𝑥→4
d) Does lim 𝑓(𝑥)
𝑥→4
exists? If so, what it is? If not, why not?
3 − 𝑥,
2) Let 𝑓(𝑥) = { 𝑥2,
,
2
𝑥<2
𝑥 = 2 be the function below
𝑥>2
a) Find lim 𝑓(𝑥) and lim− 𝑓(𝑥) , and f(2).
𝑥→2+
𝑥→2
b) Does lim 𝑓(𝑥)
𝑥→2
exists? If so, what it is? If not, why not?
c) Find lim 𝑓(𝑥) and lim− 𝑓(𝑥)
𝑥→1
𝑥→1+
d) Does lim 𝑓(𝑥)
𝑥→1
exists? If so, what it is? If not, why not?
3
3) For the function 𝑓(𝑥) = {𝑥
0
a) Graph 𝑓(x)
𝑥≠1
𝑥=1
b) Find lim− 𝑓(𝑥) and lim+ 𝑓(𝑥)
𝑥→1
𝑥→1
c) Does the lim 𝑓(𝑥)exist?
𝑥→1
2
4) For the function 𝑓(𝑥) = {1 − 𝑥
2
𝑥≠1
𝑥=1
a) Graph 𝑓(x)
b) Find lim 𝑓(𝑥)and lim+ 𝑓(𝑥)
𝑥→1
𝑥→1−
c) Does the lim 𝑓(𝑥) exist?
𝑥→1
VERTICAL ASYMPTOTE
DEFINITION VERTICAL ASYMPTOTES
A line 𝑥 = 𝑎 is a vertical asymptote of the graph of a function 𝑦 = 𝑓(𝑥) if either
𝐥𝐢𝐦 𝒇(𝒙) = ±∞
𝒙→𝒂+
𝒐𝒓 𝐥𝐢𝐦− 𝒇(𝒙) = ± ∞
𝒙→𝒂
Example 2
Discuss the behavior of the functions below
1
(𝑥 − 1)
2
𝑏) 𝑓(𝑥) =
(𝑥 + 2)2
ans:
a) 𝑓(𝑥) =
𝑛𝑒𝑎𝑟 𝑥 = 1
𝑛𝑒𝑎𝑟 𝑥 = −2
a) lim 𝑓(𝑥) = −∞ 𝑎𝑛𝑑 lim+ 𝑓(𝑥) = +∞ 𝑓(𝑥) has vertical asymptote at 𝑥 = 1
𝑥→1−
𝑥→1
b) lim 𝑓(𝑥) = +∞ 𝑎𝑛𝑑 lim+ 𝑓(𝑥) = +∞ 𝑓(𝑥) has vertical asymptote at 𝑥 = −2
𝑥→−2−
𝑥→−2
Example 3
Find the vertical asymptote for the function 𝑓(𝑥) =
1
(𝑥 − 2)2
Ans:
1
= +∞
𝑥→2 (𝑥 − 2)2
lim
y
f(x)=1/(x-2)^2
Series 1
14
12
10
8
6
4
2
x
-19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19
-2
-4
-6
1
As 𝑥 approaches 2 from the right,(𝑥−2)2 is positive and eventually becomes larger than any
preassigned number.
lim−
𝑥→2
1
(𝑥−2)2
As 𝑥 approaches 2 from the left,
assigned number.
1
= +∞
(𝑥 − 2)2
is positive and eventually becomes larger than any pre
lim+
𝑥→2
1
= +∞
(𝑥 − 2)2
Example 4
Using limit, find vertical asymptotes of the following functions:
𝑎) 𝑓(𝑥) =
𝑥2 + 1
𝑥−1
𝑏) 𝑓(𝑥) =
𝑥2 − 1
2𝑥 + 1
𝑐) 𝑓(𝑥) =
𝑥3 + 1
𝑥2
Example 5
Find the all asymptotes for the function 𝑓(𝑥) =
𝑥+3
𝑥−2
Example 6
Find the all asymptotes for the function 𝑓(𝑥) =
𝑥3 + 1
𝑥2
1.4 CALCULATING LIMIT USING THE LIMIT LAWS
BASIC LIMITS
𝒂) 𝐥𝐢𝐦 𝒌 = 𝒌
𝒙→𝒂
𝒃) 𝐥𝐢𝐦 𝒙 = 𝒂
𝒙→𝒂
𝒄) 𝐥𝐢𝐦+
𝟏
= +∞
𝒙
𝒅) 𝐥𝐢𝐦−
𝟏
= −∞
𝒙
𝒙→𝒐
𝒙→𝒐
LIMIT LAWS
Assume lim 𝑓(𝑥) = 𝐴
𝑥→𝑎
1)
and lim 𝑓(𝑥) = 𝐵
𝑥→𝑎
𝑓(𝑥) = 𝑐, then lim 𝑓(𝑥) = 𝑐
𝑥→𝑎
2)
lim 𝑐𝑓(𝑥) = 𝑐 lim 𝑓(𝑥) = 𝑐𝐴
𝑥→𝑎
𝑥→𝑎
3)
lim [𝑓(𝑥) ± 𝑔(𝑐𝑥)] = 𝐴 ± 𝐵
𝑥→𝑎
lim [𝑓(𝑥)𝑔(𝑐𝑥)] = lim 𝑓(𝑥) lim 𝑔(𝑥) = 𝐴 𝐵
4)
𝑥→𝑎
𝑥→𝑎
5)
lim
𝑥→𝑎
𝑥→𝑎
lim 𝑓(𝑥) 𝐴
𝑓(𝑥)
= 𝑥→𝑎
=
𝑔(𝑥) lim 𝑔(𝑥) 𝐵
𝑥→𝑎
𝑛
𝑛
lim √𝑓(𝑥) = 𝑛√ lim 𝑓(𝑥) = √𝐴
6)
𝑥→𝑎
𝑥→𝑎
𝑛
if √𝐴 is defined
Example 1
Using the Limit Laws, find the following limits
a) lim 5
𝑥→3
b) lim 𝑥
𝑥→𝑒
c) lim √7
𝑥→𝜋
d) lim 0
𝜋
𝑥→
5
e) lim(𝑥 3 + 4𝑥 2 − 3)
𝑥→𝑐
𝑥 4 +𝑥 2 −1
2
𝑥→𝑐 𝑥 +5
f) lim
g) lim √4𝑥 2 − 3
𝑥→2
Example 2
Find limits of rational function below
𝑥 2 + 5𝑥 + 4
𝑥→−4 𝑥 2 + 3𝑥 − 4
𝑎) lim
𝑥 2 − 4𝑥
𝑥→4 𝑥 2 − 3𝑥 − 4
𝑏) lim
𝑐) lim
9−𝑥
𝑥→9 3 −
√𝑥
Ans :
(𝑥 + 1)(𝑥 + 4)
𝑥 2 + 5𝑥 + 4
𝑥+1 3
= lim
= lim
=
2
𝑥→−4 𝑥 + 3𝑥 − 4
𝑥→−4 (𝑥 − 1)(𝑥 + 4)
𝑥→−4 𝑥 − 1
5
𝑎) lim
𝑥 2 − 4𝑥
𝑥(𝑥 − 4)
𝑥
4
= lim
= lim
=
2
𝑥→4 𝑥 − 3𝑥 − 4
𝑥→4 (𝑥 − 4)(𝑥 + 1)
𝑥→4 𝑥 + 1
5
𝑏) lim
𝑐) lim
9−𝑥
𝑥→9 3 −
√𝑥
= lim
𝑥→9
(3 − √𝑥)(3 + √𝑥)
(3 − √𝑥)
=6
Example 3
Find
4
𝑥→5 𝑥 − 7
a) lim
1
b) lim(2𝑧 − 8)3
𝑥→0
c) lim
ℎ→0 √5ℎ
5
+4+2
𝑥+3
2
𝑥→−3 𝑥 + 4𝑥 + 3
d) lim
𝑡 2 + 3𝑡 + 2
e) lim 2
𝑡→−1 𝑡 − 𝑡 − 2
√𝑥 2 + 8 − 3
𝑥+1
𝑥→−1
f) lim
Example 4
Given that lim 𝑓(𝑥) = 0 𝑎𝑛𝑑 lim 𝑔(𝑥) = −3
𝑥→4
Find
a) lim[𝑔(𝑥) + 3]
𝑥→4
b) lim [𝑔(𝑥)]2
𝑥→4
𝑐) lim 𝑥𝑓(𝑥)
𝑥→4
𝑔(𝑥)
𝑥→4 𝑓(𝑥) − 1
d) lim
𝑥→4
Example 5
Find
√𝑥 + 4 − 2
𝑥
𝑥→0
a) lim
b)
√𝑥 2 + 4 − 2
𝑥
𝑥→0
lim
Example 6 Limit of Piecewise Defined Function
1
𝑥+2
2
𝑥 < −2
Let 𝑓(𝑥) = { 𝑥 − 5 −2 < 𝑥 ≤ 3
𝑥>3
√𝑥 + 13
Find
𝑎) lim 𝑓(𝑥)
𝑥→−2
b) lim 𝑓(𝑥)
𝑥→0
c) lim 𝑓(𝑥)
𝑥→3
Example 7
𝑥−1 𝑥 ≤3
Let 𝑓(𝑥) = {
3𝑥 − 7 𝑥 > 3
Find
𝑎) lim− 𝑓(𝑥)
𝑥→3
𝑏) lim+ 𝑓(𝑥)
𝑥→3
c) lim 𝑓(𝑥)
𝑥→3
Example 8
1 + 𝑥2
Let 𝑓(𝑥) = { 2 − 𝑥
(𝑥 − 2)2
𝑥≤0
0<𝑥≤2
𝑥>2
Find
𝑎) lim 𝑓(𝑥)
𝑥→0
𝑏) lim 𝑓(𝑥)
𝑥→2
Example 9 Find limit or explain why it does not exist.
𝑥 2 − 4𝑥 + 4
a) lim 3
2
𝑥→0 𝑥 + 5𝑥 − 4𝑥
𝑥 2 − 4𝑥 + 4
b) lim 3
2
𝑥→2 𝑥 + 5𝑥 − 4𝑥
𝑥2 + 𝑥
𝑐) lim 5
𝑥→0 𝑥 + 2𝑥 4 + 𝑥 3
d) lim
𝑥→−1
e) lim
𝑥→1
𝑥2 + 𝑥
𝑥 5 + 2𝑥 4 + 𝑥 3
1 − √𝑥
1−𝑥
𝑥 2 − 𝑎2
𝑓) lim 4
𝑥→𝑎 𝑥 − 𝑎4
g) lim
𝑥→0
(𝑥 + ℎ)2 − 𝑥 2
ℎ
h) lim
ℎ→0
(𝑥 + ℎ)2 − 𝑥 2
ℎ
1
1
−2
2
+
𝑥
i) lim
𝑥
𝑥→0
1.5 LIMITS AT INFINITY AND HORIZONTAL ASYMPTOTES
Limits at infinity are used to describe the behavior of functions as the
independent variable increases or decreases without bound. If a function
approaches a numerical value L in either of these situations, write
𝐥𝐢𝐦 𝒇(𝒙) = 𝑳 𝒐𝒓
𝒙→−∞
𝐥𝐢𝐦 𝒇(𝒙) = 𝑳
𝒙→+∞
and 𝑓(𝑥) is said to have a horizontal asymptote at 𝒚 = 𝑳. A function may
have different horizontal asymptotes in each direction, have a horizontal
asymptote in one direction only, or have no horizontal asymptote.
Example 1
Find the horizontal asymptote for the function:
5𝑥 2 + 8𝑥 − 3
𝑓(𝑥) =
3𝑥 2 + 2
Ans:
lim 𝑓(𝑥) =
𝑥→−∞
5
5
𝑎𝑛𝑑 lim 𝑓(𝑥) =
𝑥→+∞
3
3
5
So the horizontal asymptote is 𝑦 = 3
Example 2: Evaluate
2𝑥 2 + 3
𝑥→+∞ 𝑥 2 − 5𝑥 − 1
lim
Factor the largest power of 𝑥 in the numerator from each term and the largest power
of 𝑥 in the denominator from each term.
You find that
2𝑥 2 + 3
𝑥→+∞ 𝑥 2 − 5𝑥 − 1
lim
3
2)
𝑥
= lim
5 1
𝑥→+∞ 2
𝑥 (1 − 𝑥 − 2 )
𝑥
𝑥 2 (2 +
2+
= lim
𝑥→+∞
3
𝑥2
5 1
1−𝑥− 2
𝑥
=
2+0
=2
1−0−0
The function has a horizontal asymptote at 𝒚 = 𝟐.
Example 3: Evaluate
𝑥3 − 2
lim
𝑥→+∞ 5𝑥 4 − 3𝑥 3 + 2𝑥
The function has a horizontal asymptote at 𝒚 = 𝟎.
Example 4: Evaluate
9𝑥 2
𝑥→+∞ 𝑥 + 2
lim
Answer:
Because this limit does not approach a real number value, the function has no horizontal
asymptote as 𝑥 increases without bound.
Limits of Polynomial as 𝒙 → ±∞
𝐥𝐢𝐦 [𝒄𝟎 + 𝒄𝟏 𝒙 + ⋯ + 𝒄𝒏 𝒙𝒏 ]= 𝐥𝐢𝐦 𝒄𝒏 𝒙𝒏
𝒙→−∞
𝒙→−∞
𝐥𝐢𝐦 [𝒄𝟎 + 𝒄𝟏 𝒙 + ⋯ + 𝒄𝒏 𝒙𝒏 ]= 𝐥𝐢𝐦 𝒄𝒏 𝒙𝒏
𝒙→+∞
𝒙→+∞
Example 4: Evaluate .
𝑎) lim 𝑥 3 − 𝑥 2 − 3𝑥
𝑥→−∞
𝑏) lim 5𝑥 7 − 𝑥 2 − 3𝑥 − 9
𝑥→−∞
𝑐) lim −4𝑥 8 − 𝑥 5 − 7𝑥 + 4
𝑥→−∞
Limit laws For Limits at Infinity
𝒏
𝐥𝐢𝐦 𝒌𝒇(𝒙) = 𝒌 𝐥𝐢𝐦 𝒇(𝒙)
𝐥𝐢𝐦 (𝒇(𝒙))𝒏 = ( 𝐥𝐢𝐦 𝒇(𝒙))
𝒙→+∞
𝒙→+∞
𝒙→+∞
𝐥𝐢𝐦 𝒌𝒇(𝒙) = 𝒌 𝐥𝐢𝐦 𝒇(𝒙)
𝒏
𝐥𝐢𝐦 (𝒇(𝒙))𝒏 = ( 𝐥𝐢𝐦 𝒇(𝒙))
𝒙→−∞
𝒙→−∞
𝒙→−∞
Infinite Limit at Infinity
𝐥𝐢𝐦 𝒇(𝒙) = ±∞
𝒙→−∞
Write limits at infinity for the graph below :
𝑖)𝑓(𝑥) = 𝑥,
𝒙→+∞
𝑖𝑖)𝑓(𝑥) = 𝑥 3 ,
𝑖𝑖𝑖) 𝑓(𝑥) = 𝑥 2
𝒙→−∞
Limit at Infinity for Rational Function
Example 5
3𝑥 5 − 2𝑥 4 + 1
𝑥→+∞ 1 − 5𝑥 2 + 3𝑥 2
lim
Ans:
3𝑥 5
= lim 𝑥 2 = +∞
𝑥→+∞ 3𝑥 3
𝑥→+∞
lim
Example 6 Limits Involving Radical
Find
3
a) lim √
𝑥→+∞
3𝑥 + 5
6𝑥 − 8
3 3𝑠 7 − 4𝑠 5
b) lim √
𝑠→+∞
2𝑠 7 + 1
Ans:
3 3𝑠 7 − 4𝑠 5
3
3𝑠 7 − 4𝑠 5 3
3𝑠 7 3 3
√ lim
√ lim
b) lim √
=
=
=√
𝑠→+∞
𝑠→+∞ 2𝑠 7 + 1
𝑠→+∞ 2𝑠 7
2𝑠 7 + 1
2
Example 7
Find
√𝑥 2 + 2
𝑥→+∞ 3𝑥 − 6
𝑎) lim
√𝑥 2 + 2
𝑥→−∞ 3𝑥 − 6
𝑏) lim
𝑐) lim
2−𝑦
𝑦→−∞ √7
+ 6𝑦 2
𝑑) lim √𝑥 6 + 5 − 𝑥 3
𝑥→+∞
𝑒) lim √𝑥 6 + 5𝑥 3 − 𝑥 3
𝑥→+∞
𝑓) lim √𝑥 2 − 3𝑥 − 𝑥
𝑥→+∞
Ans:
2−𝑦
|𝑦|
2−𝑦
1
−𝑦
𝑐) lim
= lim
=
=
𝑦→−∞ √7 + 6𝑦 2
𝑦→−∞ √7 + 6𝑦 2
7
√6
√ 2+6
𝑦
|𝑦|
2−𝑦
𝑓) lim (√𝑥 2 − 3𝑥 − 𝑥 ) ×
𝑥→+∞
=
−3
√1 − 3 + 1
𝑥
=
−3
2
(√𝑥 2 − 3𝑥 + 𝑥 )
(√𝑥 2 − 3𝑥 + 𝑥 )
=
−3𝑥
√𝑥 2 − 3𝑥 + 𝑥
=
−3𝑥
|𝑥|
2
√𝑥 −2 3𝑥 + 𝑥
𝑥
𝑥
1.5 LIMITS OF TRIGONOMETRIC FUNCTION
The trigonometric functions sine and cosine have four important limit properties:
lim
x  ( /2) 
tan x   
lim cot x  
x 0 
lim
x  ( /2) 
sec x   
lim csc x  
x 0 
lim
x  ( /2) 
tan x  
lim cot x   
x 0 
lim
x  ( /2) 
sec x  
lim csc x   
x 0 
lim
x  ( /2) 
tan x   
lim cot x  
x
lim
x  ( /2) 
sec x  
lim csc x   
x
lim
x  ( /2) 
tan x  
lim cot x   
x
lim
x   ( /2) 
sec x   
lim csc x  
x
You can use these properties to evaluate many limit problems involving the six basic trigonometric
functions.
Example 1: Evaluate
lim
𝑥→0
.
cos 𝑥
𝑠𝑖𝑛𝑥 − 3
Substituting 0 for 𝑥, you find that 𝑐𝑜𝑠 𝑥 approaches 1 and 𝑠𝑖𝑛 𝑥 − 3 approaches −3; hence,
lim
𝑥→0
cos 𝑥
1
=−
𝑠𝑖𝑛𝑥 − 3
3
Example 2: Evaluate
lim cot 𝑥
𝑥→0+
Because cot x = cos x/sin x, you find
lim+
𝑥→0
cos 𝑥
𝑠𝑖𝑛𝑥
The numerator approaches 1 and the denominator approaches 0 through positive values because
we are approaching 0 in the first quadrant; hence, the function increases without bound and
lim cot 𝑥 = +∞
𝑥→0+
and the function has a vertical asymptote at 𝑥 = 0.
Example 3: Evaluate
lim
𝑥→0
sin 4𝑥
𝑥
Multiplying the numerator and the denominator by 4 produces
Example 4: Evaluate .
lim
𝑥→0
We can start with
sec x = 1/cos x,
𝑠ec𝑥 − 1
𝑥
Example 5 : Evaluate
lim
𝑡→−3
sin(𝑡 + 3)
𝑡 3 + 3𝑡 2
Answer:
Example 6: Evaluate
lim 𝑥
𝑥→0
cos 2𝑥
sin 2𝑥
Example 7 : Evaluate
lim+
𝑥→0
sin 𝑥
|𝑥|
lim+
𝑥→0
lim−
𝑥→0
sin 𝑥
sin 𝑥
= lim+
=1
𝑥→0
|𝑥|
𝑥
sin 𝑥
sin 𝑥
= lim−
=−1
𝑥→0 −𝑥
|𝑥|
So lim+
𝑥→0
sin 𝑥
𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡
|𝑥|
Example 8
Evaluate
lim𝜋
𝑥→
2
𝑠𝑖𝑛 𝑥
𝑐𝑜𝑠 6𝑥
Example 9
Find
𝑐𝑜𝑠 𝑥
𝑥→−𝜋 𝑐𝑜𝑡 𝑥
lim
Example 10
Find the lim𝜋
𝑥→
2
sin 𝑥
cos 6𝑥
Ans:
sin 𝑥
cos 6𝑥
lim𝜋
𝑥→
2
𝜋
2 = 1 = −1
=
6𝜋 −1
𝑐𝑜𝑠 2
𝑠𝑖𝑛
Example 11
sin 3𝑥
𝑥→0
𝑥
lim
Ans:
sin 3𝑥
(3) = 3
𝑥→0 3𝑥
lim
Example 12
Find lim
𝑡→−3
𝑠𝑖𝑛(𝑡 + 3)
𝑡 3 + 3𝑡 2
Ans:
lim
𝑠𝑖𝑛(𝑡 + 3)
𝑡 3 + 3𝑡 2
lim
𝑠𝑖𝑛(𝑡 + 3)
𝑡 2 (𝑡 + 3)
𝑡→−3
𝑡→−3
1 𝑠𝑖𝑛(𝑡 + 3) 1
= (1)
𝑡→−3 𝑡 2 (𝑡 + 3)
9
lim
Example 13
Find lim 𝑥 cot 2𝑥
𝑡→0
Ans:
lim 𝑥
𝑡→0
lim
𝑥
𝑡→0 2𝑥
cos 2𝑥
cos 2𝑥
sin 2𝑥
2𝑥
1
1
= (1)(1) =
sin 2𝑥 2
2
Example 14
1 − cos 𝑥
𝑥→0
𝑥2
Find lim
Ans:
1 − cos 𝑥 1 + cos 𝑥
.
𝑥→0
𝑥2
1 + cos 𝑥
lim
1 − 𝑐𝑜𝑠 2 𝑥
𝑠𝑖𝑛2 𝑥
𝑠𝑖𝑛𝑥 2
1
1
1
=
lim
=
lim
(
)
= 12 ( ) =
2
2
𝑥→0 𝑥 (1 + cos 𝑥)
𝑥→0 𝑥 (1 + cos 𝑥)
𝑥→0
𝑥
1 + 𝑐𝑜𝑠𝑥
2
2
lim
Example 15
𝑡𝑎𝑛7𝑥
𝑥→0 𝑡𝑎𝑛2𝑥
Find lim
Example 16
5𝑥 + sin 3𝑥
𝑥→0 𝑡𝑎𝑛4𝑥 − 7𝑥𝑐𝑜𝑠2𝑥
Find lim
Ans:
1⁄
5𝑥 + sin 3𝑥
8
𝑥
.
=−
𝑥→0 𝑡𝑎𝑛4𝑥 − 7𝑥𝑐𝑜𝑠2𝑥 1⁄
3
𝑥
lim
Example 17
Find
sin ℎ
ℎ→0 1 − cos h
lim
Ans:
sin ℎ
ℎ→0 1 − cos h
lim
×
1 + cos h
1 + cos h
1 + cos ℎ 2
=
ℎ→0
sin ℎ
0
= lim
𝐷𝑁𝐸
Example 18
Find
1 − cos 3ℎ
ℎ→0 𝑐𝑜𝑠 2 5ℎ − 1
lim
Ans:
1 − cos 3ℎ
1 + cos 3ℎ 𝑐𝑜𝑠 2 5ℎ + 1 −9
×
×
=
ℎ→0 𝑐𝑜𝑠 2 5ℎ − 1
𝑐𝑜𝑠 2 5ℎ + 1 1 + cos 3ℎ
50
lim
Example 19
Find
𝑠𝑖𝑛𝜋𝑥
𝑥→1 𝑥 − 1
lim
Ans:
𝐿𝑒𝑡 ℎ = 𝑥 − 1, 𝑥 → 1, ℎ → 0
𝑠𝑖𝑛𝜋𝑥
𝑥→1 𝑥 − 1
𝑠𝑖𝑛𝜋(ℎ + 1)
sin 𝜋𝑡𝑐𝑜𝑠 𝜋 + 𝑐𝑜𝑠𝜋𝑠𝑖𝑛𝜋
𝑠𝑖𝑛𝜋𝑡
𝑠𝑖𝑛𝜋𝑡
lim
= lim
= − lim
= − lim
𝜋 = −𝜋
ℎ→0
ℎ→0
ℎ→0
ℎ→0
ℎ
ℎ
𝑡
𝜋𝑡
lim
Example 20
𝑐𝑜𝑠 𝑥
1
= −
x→0 𝑠𝑖𝑛𝑥 − 3
3
Find lim
Example 21
Find lim+ 𝑐𝑜𝑠 𝑥 = lim+
x→0
x→0
𝑐𝑜𝑠 𝑥
= −∞
𝑠𝑖𝑛𝑥
1.6 CONTINUITY
A function is said to be continuous on an interval if its graph has no breaks, jumps or holes in that
interval.
Example
𝑎) 𝑓(𝑥) = 3𝑥 3 + 2𝑥 + 1
𝑏) 𝑓(𝑥) =
1
𝑥
Definition of Continuity
The function 𝑓 is continuous at 𝑥 = 𝑐 if f is defined at 𝑥 = 𝑐 and
lim 𝑓(𝑥) = 𝑓(𝑐)
𝑥→𝑐
The function is continuous on an interval (𝑎, 𝑏) if it is continuous at every point in the interval.
Example 1: Are the functions continuous on the given interval?
𝑎) 𝑓(𝑥) = 𝑥 + 2
on −3 ≤ 𝑥 ≤ 3
𝑏) 𝑓(𝑥) = 2𝑥
on
0 ≤ 𝑥 ≤ 10
𝑐) 𝑓(𝑥) = 𝑥 2 + 2
on
0≤𝑥≤5
on
2≤𝑥≤3
on
0≤𝑥≤2
1
𝑥−1
1
𝑒) 𝑓(𝑥) =
𝑥−1
𝑑) 𝑓(𝑥) =
Example 2: Determine whether the following function are continuous at 𝑥 = 2
𝑎) 𝑓(𝑥) =
𝑥2 − 4
𝑥−2
𝑥2 − 4
𝑏) 𝑔(𝑥) = { 𝑥 − 2 𝑥 ≠ 2}
3
𝑥=2
𝑥2 − 4
𝑐) 𝑔(𝑥) = { 𝑥 − 2 𝑥 ≠ 2}
4
𝑥=2
Definition
If a function 𝒇 is continuous at 𝒙 = 𝒂 then we must have the following three
conditions.
1. 𝒇(𝒂) is defined; in other words, 𝒂 is in the domain of 𝒇.
Example 1
The function
𝑥2 − 9
𝑓(𝑥) =
𝑥−3
fails to be continuous at 𝒙 = 𝟑 since 𝟑 is not in the domain of 𝒇.
Condition 1 is not true.
Example 2
The function
1
𝑖𝑓
𝑔(𝑥) = {𝑥
2 𝑖𝑓
fails to be continuous at 𝒙 = 𝟎 since the limit
𝑥≠0
𝑥=0
lim 𝑔(𝑥) does not exist, therefore Condition 2 is not true.
𝑥→0
Example 3
The function
𝑥2 − 9
ℎ(𝑥) = { 𝑥 − 3
7
𝑖𝑓
𝑥≠3
𝑖𝑓
𝑥=3
does satisfy condition 1 since 3 is in the domain of 𝒉, 𝒉(𝟑) = 𝟕, and does satisfy condition 2
since
(𝑥 − 3)(𝑥 + 3)
𝑥2 − 9
= lim
= lim 𝑥 + 3 = 6
𝑥→3 𝑥 − 3
𝑥→3
𝑥→3
𝑥−3
lim
does exist.
However, since these two numbers are different, condition 3 is violated and 𝒉(𝒙) fails to be
continuous at 𝒙 = 𝟑.
Example 4:
Determine whether the function below is continuous.
𝑥3 + 2
𝑥 ≤ −1
2
𝑓(𝑥) = {𝑥 + 𝑥 + 1 −1 < 𝑥 < 1}
𝑥4 + 1
𝑥≥1
Ans:
𝑥 is continuous at 𝑥 = −1
𝑥 is discontinuous at 𝑥 = 1
Example 5:
2𝑥 3 + 16
Given 𝑓(𝑥) = {𝑥 2 + 𝑏𝑥 + 𝑐
3𝑥 4 − 48
𝑥 ≤ −2
−2 < 𝑥 < 2}
𝑥≥2
Find 𝒃 and 𝒄 such that 𝒇(𝒙) is continuous everywhere
Ans: 𝑏 = 0, 𝑐 = −4
TUTORIAL
1.2 CALCULATING LIMIT USING LIMIT LAW
√𝑥 − 3
𝑥→9 𝑥 − 9
1) Find lim
2) Find
lim
𝑥→4
4𝑥 − 𝑥 2
2 − √𝑥
√𝑥 2 + 12 − 4
𝑥→2
𝑥−2
3)Find lim
𝑢4 − 1
𝑢→1 𝑢 3 − 1
4)Find lim
5) 𝑆uppose the lim 𝑓(𝑥) = 1 and lim 𝑔(𝑥) = −5 , show that lim
𝑥→0
𝑥→0
𝑥→0
2 𝑓(𝑥) − 𝑔(𝑥)
(𝑓(𝑥) +
2
7)3
=
7
4
𝑣3 − 8
𝑣→2 𝑣 4 − 16
6) Find lim
5
√5ℎ(𝑥)
=
𝑥→1 𝑝(𝑥)(4 − 𝑟(𝑥))
2
7) Suppose lim ℎ(𝑥) = 5and lim 𝑝(𝑥) = 1 , show that lim
𝑥→1
𝑥→1
8) Suppose
lim 𝑓(𝑥) = 7 and lim 𝑔(𝑥) = −3 , find
𝑥→𝑏
𝑥→𝑏
a) lim (𝑓(𝑥) + 𝑔(𝑥))
𝑥→𝑏
b) lim 𝑓(𝑥)𝑔(𝑥)
𝑥→𝑏
c) lim 4𝑔(𝑥)
𝑥→𝑏
𝑓(𝑥)
𝑥→𝑏 𝑔(𝑥)
d) lim
9) If lim
𝑥→4
𝑓(𝑥) − 5
= 1, find lim 𝑓(𝑥)
𝑥→4
𝑥−2
10) Suppose lim 𝑝(𝑥) = 4 , lim 𝑟(𝑥) = 0 and lim 𝑠(𝑥) = −3 , find
𝑥→−2
a) lim
𝑥→−2
𝑥→−2
(𝑝(𝑥) + 𝑟(𝑥) + 𝑠(𝑥))
b) lim
𝑝(𝑥)𝑟(𝑥)𝑠(𝑥)
c) lim
4𝑠(𝑥)
d) lim
(−4𝑝(𝑥) + 5𝑟(𝑥))
𝑠(𝑥)
𝑥→−2
𝑥→−2
𝑥→−2
𝑥→−2
𝑓(𝑥)
= 1, find
2
𝑥→−2 𝑥
11) If lim
,
𝑎)
𝑏)
lim 𝑓(𝑥)
𝑥→−2
lim
𝑥→−2
𝑓(𝑥)
𝑥
12) If
𝑓(𝑥)
= 1, find
2
𝑥→0 𝑥
lim
a)
lim 𝑓(𝑥)
𝑥→0
𝑓(𝑥)
𝑥→0 𝑥
b) lim
13)
𝑎) If
𝑓(𝑥) − 5
= 3, find lim 𝑓(𝑥)
𝑥→2 𝑥 − 2
𝑥→2
lim
𝑓(𝑥) − 5
= 4, 𝑓𝑖𝑛𝑑 lim 𝑓(𝑥)
𝑥→2 𝑥 − 2
𝑥→2
𝑏) If lim
Ans:
1) 1/6
2)
3) ½
4)
5) 7/4
1.3 INFINITE LIMIT AND VERTICAL ASYMPTOTES
Using limits , identify vertical asymptote for the function below:
𝑎) 𝑓(𝑥) =
𝑥2 − 2
𝑥−2
𝑥3 − 𝑥 + 3
𝑏) 𝑔(𝑥) =
𝑥
𝑐) 𝑓(𝑥) =
𝑑) ℎ(𝑥) =
−𝑥 3 + 3𝑥 2 + 𝑥 − 1
𝑥−3
𝑥5 − 𝑥3 + 3
𝑥2 − 1
1.4 FIND LIMITS AT INFINITY
3𝑥 + 1
𝑥→+∞ 2𝑥 − 5
6) lim
3𝑥 + 7
𝑥→−∞ 𝑥 2 − 2
7) lim
𝑥−2
𝑥→−∞ 𝑥 2 + 2𝑥 + 1
8) lim
3 2 + 3𝑥 − 5𝑥 2
9) lim √
𝑥→−∞
1 + 8𝑥 2
11) lim
2−𝑦
𝑦→−∞ √7 +
6𝑦 2
√5𝑥 2 − 2
𝑥→−∞ 𝑥 + 3
12) lim
6 − 𝑡3
𝑡→−∞ 7𝑡 3 + 3
13) lim
√3𝑥 4 + 𝑥
𝑥→−∞ 𝑥 2 − 8
14) lim
15) lim (√𝑥 2 − 𝑥 − 𝑥)
𝑥→+∞
𝑥 + 4𝑥 3
𝑥→−∞ 1 − 𝑥 2 + 7𝑥 3
16) lim
17) lim (√𝑥 2 + 𝑎𝑥 − 𝑥)
𝑥→+∞
18) lim (√𝑥 2 + 𝑎𝑥 − √𝑥 2 + 𝑏𝑥)
𝑥→+∞
19) lim (√𝑥 2 − 3𝑥 − 𝑥)
𝑥→+∞
Ans:
1) (a)2,1 (b) no lim+ 𝑓(𝑥) ≠ lim− 𝑓(𝑥)
𝑥→2
3) (a)graph (b)1,1 (c)yes, 1
𝑥→2
(c)3,3
(d)yes 3
5) 𝑎) √3, c) 1, e)
2
, g) 1, i) √2
√5
6)3/2
7)0
3
9)− √5
1
√6
11)
13)−1/7
15)0
17)𝑎/2
1.5 LIMIT OF TRIGONOMETRY FUNCTION
𝜃2
𝜃→0 1 − 𝑐𝑜𝑠𝜃
1) lim
1 − 𝑐𝑜𝑠3ℎ
2
ℎ→0 𝑐𝑜𝑠 5ℎ − 1
2) lim
2 − 𝑐𝑜𝑠3𝑥 − 𝑐𝑜𝑠4𝑥
𝑥
𝑥→0
3) lim
𝑡𝑎𝑛3𝑥 2 + 𝑠𝑖𝑛2 5𝑥
4) lim
𝑥2
𝑥→0
tan 𝑎𝑥
, 𝑎 ≠ 0, 𝑏 ≠ 0
𝑥→0 sin 𝑎𝑥
5) lim
𝑥 2 − 3𝑠𝑖𝑛𝑥
𝑥
𝑥→0
6) lim
𝜋−𝑥
;
𝑥→𝜋 𝑠𝑖𝑛𝑥
7) lim
𝐻𝑖𝑛𝑡: 𝐿𝑒𝑡 𝑡 =
1
1
8) lim 𝑥 (1 − 𝑐𝑜𝑠 ) ; 𝑡 =
𝑥→−∞
𝑥
𝑥
sin(𝜋𝑥)
𝑥→1 𝑥 − 1
9) lim
𝜋 𝜋
−
2 𝑥
10) lim𝜋
𝑥→
Ans:
1) 2
2)
3) 0
4)
5) a/b
6)
7) 1
8)
9) –𝜋
10)
4
𝑡𝑎𝑛𝑥 − 1
𝑥 − 𝜋⁄4
1.6 CONTINUITY
1. Graph the function, then discuss limits, one sided limits, and continuity of 𝑓
at 𝑥 = −1,0 𝑎𝑛𝑑 1.
1
−𝑥
𝑎) 𝑓(𝑥) = 1
−𝑥
{1
0
1
𝑏) 𝑓(𝑥) = 𝑥
0
{1
𝑖𝑓
𝑖𝑓
𝑖𝑓
𝑖𝑓
𝑖𝑓
𝑥 ≤ −1
−1 < 𝑥 < 0
𝑥=0
0<𝑥<1
𝑥≥1
𝑖𝑓
𝑥 ≤ −1
𝑖𝑓
0 < |𝑥| < 1
𝑖𝑓
𝑖𝑓
𝑥=1
𝑥>1
2. At which points do the functions in exercises continuous.
𝑎) 𝑦 =
1
− 3𝑥
𝑥−2
𝑎𝑛𝑠: 𝑒𝑥𝑐𝑒𝑝𝑡 𝑎𝑡 𝑥 = 2
𝑏) 𝑦 =
1
+4
(𝑥 + 2)2
𝑐) 𝑦 =
𝑥2
𝑥+1
− 4𝑥 + 3
𝑎𝑛𝑠: 𝑒𝑥𝑐𝑒𝑝𝑡 𝑥 = 3
𝑥2
𝑥+3
− 3𝑥 − 1
𝑎𝑛𝑠: 𝑒𝑥𝑐𝑒𝑝𝑡 𝑥 = 5 , 𝑥 = 2
𝑑) 𝑦 =
𝑒)
𝑦=
1
𝑥2
−
|𝑥| + 1 2
𝑓) 𝑦 = √2𝑥 + 3
4
𝑔) 𝑦 = √3𝑥 − 1
𝑎𝑛𝑠: 𝑒𝑥𝑐𝑒𝑝𝑡 𝑎𝑡 𝑥 = 2
ans: continuous everywhere
𝑎𝑛𝑠:
all x ≥ −3/2
𝑎𝑛𝑠: 𝑥 ≥
1
3
1⁄
3
ℎ) 𝑦 = (2𝑥 − 1)
𝑎𝑛𝑠: 𝑎𝑙𝑙 𝑥
1⁄
5
𝑖) 𝑦 = (2 − 𝑥)
𝑎𝑛𝑠: 𝑎𝑙𝑙 𝑥
3. For what value of a is 𝑓(𝑥) = {
𝑥2 − 1
2𝑎𝑥
4. For what value of a is 𝑓(𝑥) = {
𝑥
𝑏𝑥 2
𝑖𝑓
𝑖𝑓
𝑖𝑓
𝑖𝑓
𝑥<3
, continuous at every 𝑥?
𝑥≥3
Ans: 4/3
𝑥 < −2
, continuous at every 𝑥?
𝑥 ≥ −2
Ans: -1/2
5. Discuss limits, one sided limits, and continuity of 𝑓
at 𝑥 = 0, 1 𝑎𝑛𝑑 2.
11
f(x)=(x^2)-1
f(x)=2x
f(x)=-2x+4
10
f(x)=0
Series 1
9
8
7
6
5
4
3
2
1
x
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
-1
-2
-3
2
3
4
5
6
7
8
9
10
11
ADDITIONAL QUESTIONS
Limits and Continuity
1.
Find the limits:-
(𝑎)
(𝑏)
−3x 4 − 1
lim
x → −∞ 𝑥 2 (𝑥 − 1)
√x 2 (√2(x 2 + 4)
lim
(
)
𝑥→2
3x 2 − 4
Ans: +∞ , 1
2.
Given 𝑓(𝑥) is a function defined by
𝑥≤1
𝑥𝑥 2 + a
𝑓(𝑥) = {𝑚𝑚𝑥 + 𝑐 1 < 𝑥 ≤ 3
𝑥>3
2𝑥 2
(a)
(b)
Determine the value of 𝑎 if
Evaluate 𝑚 and 𝑐 if
lim
𝑓(𝑥) = 6 .
𝑥 → 1−
lim
lim
𝑓(𝑥) and
𝑓(𝑥) exist.
𝑥→1
𝑥→3
Ans: 𝒂 = 𝟓, 𝒎 = 𝟔, 𝒄 = 𝟎
3.
Evaluate the limit.
lim 1 − cos 7k
𝑘 → 0 cos 5𝑘 − 1
Ans: −
𝟒𝟗
𝟐𝟓
4. Evaluate the limit
lim
3𝑥 + 1
𝑥→−∞ √𝑥 2
+1
Ans: 3
5. Find the value of the following limit:
lim−
𝑥→1
√2𝑥 (𝑥 − 1)
|𝑥 − 1|
Ans: −√𝟐
6. Find the value of k, if possible, that will make the function below
continuous everywhere.
𝑥 + 2𝑘,
𝑥≤1
𝑓(𝑥) = {
𝑘𝑥 2 + 𝑥 + 1, 𝑥 > 1
Ans: k = 1
7. Find values of x at which f is not continuous

3  x, 0  x  2

 4x 1
f ( x)  
, 2 x4
 x 1
 15
 7  x , 4  x  7
Ans:
𝒅𝒊𝒔𝒄𝒐𝒏𝒕𝒊𝒏𝒐𝒖𝒔 𝒂𝒕 𝒙 = 𝟐 𝒂𝒏𝒅 𝒙 = 𝟒
8. Find the interval at which
𝑓(𝑥) =
(𝑥 − 1)2
𝑥2 − 1
is continuous.
Ans: (−∞, −𝟏) ∪ (−𝟏, 𝟏) ∪ (𝟏, +∞)
9 . Evaluate the limit if exist:
lim𝜋
𝑥→
3
𝑠𝑖𝑛(𝑥 − 𝜋3)
2
(𝑥 − 𝜋3)
Ans: does not exist
10. Given f x  
3 x
x2  9
. Evaluate:
(a) lim  f x 
{ans: 1/6}
(b) lim  f x 
{ans: -1/6}
(c) lim f x 
{ans: DNE}
x3
x3
x3
11. lim
x 1
x 1
1  x 

x 1
{ans: -1/4}
2  x  x2
does not exist.
x  2 4  2x  2x 2
12. Explain why lim
13
lim
x  
x 2  2x  1  x 2  7x  3
{ans: 5/2}
2 x 2  a ,
x 1

14. If f  x    bx  c, 1  x  3
 x3 ,
x3

(a) Determine the value of a if lim  f x   5.
{ans: 7}
(b) Evaluate b and c if lim f x  and lim f x  exist.
{ans: b=16, c=-21}
x 1
x 1
x3
15. Determine whether f is continuous at x = 1 if
 x 1
 x 1 , x  1

f x     1,
x 1
2  3 x , x  1


{ans: continuous}
16.
 x2 ,
0 x2

f  x    x  2, 2  x  4
2ax  7,
x4

Given
(a) Determine whether lim f x  exist.
x2
{ans: exists}
(b) Is f continuous at x = 2? Why?
{ans: not continuous}
(c) What is the value of a if f is continuous at x = 4?
{ans: 13/8}
 

x
{ans:  }
17. Evaluate: lim x sin 
x  
18. Evaluate: lim
t0
19. Evaluate:
20. Given
Evaluate:
cos t  1
t2
lim1
𝑥→
√2
{ans: -1/2}
4𝑥 2 − 2
8𝑥 3
− √8
|2 + 3𝑥 − 2𝑥 2 |
𝑓(𝑥) =
+ 2𝑥 − 1
2𝑥 − 4
{ 𝑎𝑛𝑠:
√2
}
4
𝑥 →2+
(b)
𝑥 →2−
(c)
21. Evaluate:
11
}
2
1
{ 𝑎𝑛𝑠: }
2
{ 𝑎𝑛𝑠: 𝐷𝑁𝐸 }
lim 𝑓(𝑥)
(a)
{ 𝑎𝑛𝑠:
lim 𝑓(𝑥)
lim 𝑓(𝑥)
𝑥→2
lim
𝑥→0
√𝑥 + 𝑘 − √𝑘
2 − √𝑥 + 4
{ 𝑎𝑛𝑠: −
𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
2−𝑡
is +∞, −∞, or does not exist.
𝑡 → −2 𝑡 2 − 4
Give reason for your answer.
22. Determine whether
23. Evaluate:
2
√𝑘
}
lim
lim √2𝑥 2 − √2𝑥 2 − 6𝑥
{ 𝑎𝑛𝑠: 𝐷𝑁𝐸 }
{ 𝑎𝑛𝑠:
𝑥 → +∞
3
√2
}
24. Given:
𝑐𝑥 3 − 8,
𝑓(𝑥) = {𝑑,
2𝑥 2 + 𝑐,
if lim 𝑓(𝑥) exists, find the value of 𝑐.
𝑥<3
𝑥=3
𝑥>3
𝑥→3
Hence, if f is continuous at x = 3, find d.
{ 𝑎𝑛𝑠: 𝑐 = 1, 𝑑 = 19 }
25. Given:
𝑥 2 + 3𝑥 − 10
,
𝑥<2
|𝑥 − 2|
𝑓(𝑥) = 𝑝,
𝑥=2
5𝑥 − 𝑞,
2<𝑥<3
{ 𝑟,
𝑥>3
(a) If f is continuous at x =2, find p and q.
(b) Find r if f is discontinuous at x = 3.
{ 𝑎𝑛𝑠: 𝑝 = −7, 𝑞 = 17 }
{ 𝑎𝑛𝑠: 𝑟 ≠ −2 }
26. Determine the intervals of x where h is continuous.
√𝑥 2 − 9
ℎ(𝑥) =
(3 − 𝑥)(𝑥 + 5)
{ 𝑎𝑛𝑠: (−∞, −5) ∪ (−5, −3] ∪ (3, +∞) }
27. Evaluate: lim
tan 8𝜃
sin 4𝜃
28. Evaluate:
cos 𝑥 − 1
𝑥2
𝜃→0
lim
𝑥→0
{ 𝑎𝑛𝑠: 2 }
1
{ 𝑎𝑛𝑠: − }
2