-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
i
MA 125- Calculus
Table of Contents Page
1.Definition of Limits
Left hand and right hand limit
Calculating the limit at a point.
2..Exercise # 1
4 Limit Theorems
5 Solution to Exercise # 1
6 Exercise # 2
7.Finding Limits of Functions
In Class Assessment
9 Exercise # 3-4
10 Solution-Exercise # 4
11 Exercise #5
12Continuity
14Worksheet # 1
15 Worksheet # 2
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
1
Module 1.2- Limits Definition The limit of a function f(x), is
the value that f(x) approaches, as x tends towards a. This is
written as ax
lim f(x) = L. This means that as x gets closer to a, from either
side (but not
equal to a) the value of f(x) gets closer to L.
Suppose that our function f(x) is x2
+ 2. If one needs to find the value of the limit of f(x) at
x equal 2. We need to find out the value of f(x) as x approaches
2 from above and below.
This is easily done by observing a table of values.
x 1.70 1.75 1.80 1.85 1.90 1.95 1.99
f(x) 4.89 5.0625 5.24 5.4225 5.61 5.8025 5.9601
Notice for this table, 2 is approached from the left hand side.
If 2, is approached from the
right hand side the following table illustrates the result.
x 2.30 2.25 2.20 2.15 2.10 2.05 2.01
y 7.29 7.0625 6.84 6.6225 6.41 6.2025 6.0401
From the tables above, one should realize that as x approaches 2
the value of x2 +2 tends
towards 6.
Left hand limit & Right hand limit.
The left hand limit is the value that f(x) approaches as one
tends to where x = a from the
left hand side. From the example above, we approached 2 from the
left, this may be
written as )2(lim 22
xx
= 6. If two was approached from the right hand side it would
be
expressed as, 6)2(lim 22
xx
.
Please note that the left hand and right hand limit are equal.
The limit only exists if the
left hand and right hand limit are equal.
Lets investigate for f(x) = 2
232 2
x
xx. By using the table of values below one may be
able to investigate for 2
limx 2
232 2
x
xx.
x 1.75 1.80 1.85 1.90 1.95 1.99
y 4.5 4.6 4.7 4.8 4.9 4.98
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
2
x 2.25 2.20 2.15 2.10 2.01
y 5.52 5.4 5.3 5.2 5.02
By examining the tables it can be seen that it is true to say
that as x tends to 2 the value of
f(x) approaches 5. Hence, the limit as x tends to 2 is 5. This
may be written as
2limx
2
232 2
x
xx= 5.
This method of calculating limits is tedious; hence the
alternative is to calculate limit the
limit at the respective points. The method of calculation used,
is dependent on the type of
function and also the value that the limit tends to.
In this section we will discuss the most suitable method for
calculating the limits of some
functions.
Find the limits of the following functions.
1. )23(lim
2
x
x. The limit of this function may be determined by direct
substitution.
)23(lim2
xx
= 3(2)-2 =6-2 = 4
2. )42(lim 22
xxx
By direct substitution, implies 2(2)2 -4(2)= 8 -8 = 0
N.B. Direct substitution is used for functions that are
determined by
polynomials.
3. 1
lim2
1
x
xx
x, Notice that if direct substitution is used with this
question, the result
will be undefined. This solution is not true. The real solution
is determined by
factorization, )1(
)1(lim
1
x
xx
x= x
x 1lim
.
Notice that this rational expression is simplified to a linear
function, xx 1lim
=1.
This strategy is employed when the direct substitution method
fails especially
when the function is a rational function.
4. 2
3lim
22 xx By direct substitution
2
3lim
22 xx =
22
32
=6
3
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
3
This question posed little difficulty since its not a function
of the format )(
)(
xg
xf.
In other words it isnt a rational function so direct
substitution could be used to evaluate the limit.
5. 3
34lim
2
3
x
xx
x.
For this question direct substitution fails again. The rational
function may be
reduced by using factorization to simplify the expression.
3
34lim
2
3
x
xx
x =
)3(
)1)(3(lim
3
x
xx
x= )1(lim
3
x
x= -3 +1= -2
Exercise 1.
Compute the limits of the functions.
1. )61(lim
1x
x
2. x
x
x
5
1lim
2
5
3. x
xx
x
3lim
2
0
4. 1
1lim
2
1
x
x
x
5. 2
42lim
2
2
x
xx
x
6. 3
6lim
2
3
x
xx
x
7. x
x
x
4
16lim
2
4
8. 25
102lim
25
x
x
x
9. 65
6lim
2
2
6
xx
xx
x
10. )11392(lim 2352
xxxx
11. )73(lim 22
xxx
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
4
12. )72)(5(lim
)45(lim
3
23
0
xx
tt
x
x
13. 3 22
23lim
xxx
14. 3
9lim
2
3
x
x
x
15. x
x
x
1)1(lim
2
0
16. 9lim 24
xx
Limit Theorems
If the limit of f(x) and g(x) exist, that is if Mxfax
)(lim and .)(lim Nxgax
I. ))()((lim xgxfax
= )(lim xfax
+ )(lim xgax
= M +N
II. )(lim xcfax
=c )(lim xfax
=cM
III. )()(lim xgxfax
= )(lim)(lim xgxfaxax
= MN
IV. )(
)(lim
xg
xf
ax=
)(lim
)(lim
xg
xf
ax
ax
= N
M. This is true only if )(lim xg
ax 0
V. ccax
lim
VI. 2))((lim xfax
= 2))(lim( xfax
= M2
Please note that there exist other theorems of limits that the
reader will find useful.
Limits as one Approaches Infinity
Calculate 1
1lim
x
x
x.
If one uses direct substitution, it would result in a large
numerator and a large
denominator. It is not mathematically sound to say infinity
divided by infinity gives 1.
Hence, the solution is accomplished by dividing each term within
the numerator and
the denominator by the highest power of the variable within the
denominator.
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
5
Hence the result will be,
xx
xxx
x
x 1
1
lim
. This is simplified to,
x
xx 1
1
11
lim
.
Please note that x
1 approaches zero as x tends towards infinity hence the limit of
the
function is 1.
Eg. 2. Calculate 2
53lim
2
2
x
x
x
22
2
22
2
2
53
lim
xx
x
xx
x
x
=
xlim
2
2
21
53
x
x
,
By substitution the result will be 3 since dividing by infinity
result is zero.
Solution to Exercise 1.
1. 1-6= -5
2.
0
26
55
152
3. 3)3(lim)3(
lim00
x
x
xx
xx
4. 2)1(lim)1(
)1)(1(lim
11
x
x
xx
xx
5. 42lim)2(
)2(2lim
22
x
x
xx
xx
6. 5)2(lim)3(
)2)(3(lim
33
x
x
xx
xx
7. 8)4(lim)4(
)4)(4(lim
)4(
)4)(4(lim
444
x
x
xx
x
xx
xxx
8. 5
1
)5(
2lim
)5)(5(
)5(2lim
55
xxx
x
xx
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
6
9.7
6
1lim
)1)(6(
)6(lim
66
x
x
xx
xx
xx
10.64-72 + 12-11= -7
11. 4 +6 -7 = 3
12.-8
13. 28264 33
14. 6)3(lim)3(
)3)(3(lim
33
x
x
xx
xx
15. 22lim2
lim121
lim0
2
0
2
0
x
x
xx
x
xx
xxx
16. 59lim 24
xx
Exercise 2.
Evaluate the following limits.
1. x
limx
x
21
2. 9
62lim
2
x
xx
x
3. x
x
x 21lim
4. 2
1lim
xx
5. 2
12lim
x
x
x
6. 1
121512lim
2
2
t
tt
x
7. 13
68lim
x
x
x
8. 13
23lim
2
x
x
x
9. 34
23lim
2
x
x
x
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
7
10. 123
45lim
2
32
xx
xx
x
Finding Limits by Rationalizing
Evaluate the limit of h
h 11 as h tends to zero.
Method - To rationalize the expression, both numerator and the
denominator must be
multiplied by the conjugate of the numerator. The conjugate for
this example is
h1 +1.
0limh h
h 11 x
11
11
h
h=
)11(
11
hh
h=
)11( hh
h
=0
limh 11
1
h =
2
1
Attempt
7
32lim
7
x
x
x N. B. The conjugate of the numerator is 2x +3
Answer = 6
1
In Class Assessment Calculus
1. Given that )(lim xfax
=-3, )(lim xgax
= 0 & )(lim xhax
= 8.
Find the limits that exist. If the limit does not exist explain
why.
a) ax
lim [f(x) +h(x)]
b) ax
lim [f(x)]2
c) ax
lim 3 )(xh
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
8
d) )(
)(lim
xh
xf
ax
e) )(
)(lim
xh
xg
ax
f) )(
)(lim
xg
xf
ax
g) )()(
)(2lim
xfxh
xf
ax
2. Evaluate the following limits.
a) 4
limx
(5x2 -2x +3)
b) 2
limt
(t +1)9(t2 -1)
c) 2
limuu4 +3u + 6
d)
372
9lim
2
2
3
tt
t
t
e) h
h
h
8)2(lim
3
0
3. Show that
12
54lim
2
2
x
xx
x = 2
4. Evaluate the following limits
a) 4
53lim
x
x
x
b)
42
5lim
3
3
xx
xx
x
c)
1
2lim
23
2
tt
t
t
5. Sketch a graph that illustrate the following
characteristics
)(lim0
xfx
,
)(lim0
xfx
, 1)(lim
xfx
, 1)(lim
xfx
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
9
Exercise 3.
Calculus
Answer all questions showing appropriate working
1. Given that ax
lim f(x) = -3, ax
lim g(x) = 0 and ax
lim h(x) = 8
Find the limits of
(a) ax
lim [f(x) + h(x)] (b) ax
lim [f(x)]2
(c) ax
lim )(
)(
xg
xf (d)
axlim
)(
)(
xh
xf
2. Evaluate the limits of the following using the laws of limits
where appropriate
(a) 4
limx
(5x2 -2x + 3) (b)
2lim
x 46
122
xx
x
(c) 3
limt 372
9
tt
t (d)
2limx
2
62
x
xx
(f) x
lim 4
53
x
x
Exercise 4.
1. Suppose 3)(lim2
xfx
and 4)(lim2
xgx
. Use the limit rules to find the following
limits.
a. )(5)(lim2
xgxfx
b. )()(2lim2
xgxfx
c. )(lim
)(3lim
2
2
xf
xg
x
x
Find the limits
2. 1
1lim
2
3
x
xx
x
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
10
3. 2
6lim
2
2
x
xx
x
4. 4
2lim
4
x
x
x
5. 32
1 1
12lim
x
xx
x
Solution -Exercise 4
6. Suppose 3)(lim2
xfx
and 4)(lim2
xgx
. Use the limit rules to find the following
limits.
a. 23)4(53)(5lim)(lim)(5)(lim222
xgxfxgxfxxx
b. 2432)(lim)(2lim)()(2lim222
xgxfxgxfxxx
c.
43
43
)(lim
)(3lim
2
2
xf
xg
x
x
Find the limits
7. 2
5
4
5
1lim
133
1lim
)1(lim
1
1lim
3
3
3
2
3
2
3
xx
xx
x
xx
xx
x
x
8.
532)3(lim)2(lim
)3)(2(lim
)2(lim
6lim
2
6lim
2
2
2
2
2
2
2
2
x
x
xx
x
xx
x
xx
x
x
x
x
x
x
9.
4
1
22
1
24
1
2
1lim
24
4
24
2
2
2
4lim
2lim
4
2lim
4
22
4
4
4
x
xx
x
xx
x
x
x
x
x
x
x
x
x
x
x
10.
existnotdoes
xxxx
xx
x
xx
x
x
x
xlim
0
1
11
1
1
1lim
111lim
11lim
1
12lim
1
1
1
3
2
1
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
11
Exercise 5.
1. Evaluate the following limits
(a) 2
38
lim4
x
xx
(b) 4
16lim
2
4
x
x
x
(c) 2
65lim
2
2
x
xx
x
(d) 43
32lim
2
2
1
xx
xx
x
(e) 3
9lim
9
x
x
x
2. Determine the values of x for which the following functions
are
discontinuous.
(a) f(x) = 2
232
x
xx
(b) f(x) = 86
1032
2
xx
xx
3. Given that 3
limx
{f(x) + 3x} =1,evaluate )}(9{lim3
xfx
4. Given that 2
limx
{4f(x)} = 5,evaluate 2
limx
{f(x) + 2x}
5. Find the limits of the following functions
a) 10
43lim
x
x
x
b) 12
63lim
2
x
xx
x
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
12
Continuity
A function is continuous at a point where x = c if the following
conditions are satisfied.
f(c) is defined
cx
lim f (x) exist
cx
lim f (x) = f(c)
If the function is not continuous it is said to be
discontinuous. All functions
which are defined by polynomials are continuous inclusive of any
exponential
functions and the log function y = logax for all x> 0.
N.B. When a function is defined by a given formula it is usually
continuous for all
values within the domain. If a function has a change in the
defining formula then
there exists some value of x, within the domain at which the
function is
discontinuous.
Points of Discontinuity
A rational functions point of discontinuity may be easily
identified since
this is occurs whenever the denominator of the function is zero.
The function is
therefore undefined at this point. A function may also have more
than one point of
discontinuity. Hence, one has to explore all the possibilities
for the function to be
discontinuous.
In this section we examine two types of rational functions and a
piecewise
function in order to determine their points of
discontinuity.
Example 1. f(x) = 2
5
x
x and g(x) =
)2(
5
xx
x.
Notice that f(x) is discontinuous at x = 2 while g(x) is
discontinuous at x = 0 and
x = 2. For these values of x within the domain the function is
undefined.
Example 2. Given the function f(x) = 1 if x < 2
x +3 if 2 x 4 7 if x > 4
This is a piecewise function because it has three different
expressions by which it
is defined and each expression uses a specified domain. To check
whether the
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
13
function is discontinuous, a point within the domain has to be
identified where
this occurs. A sketch of this graph will be useful to identify
the point where
discontinuity occurs. Make an attempt to draw the graph.
From the sketch, the graph should indicate that f(x) is
discontinuous at x = 2. Also
by observations and calculations when x < 2, f(x) =1, but at
x = 2, f(x) = 2 +3 =5.
Since there is a jump from f(x) =1 to f(x) = 5, this is a jump
discontinuity, f(x) is
discontinuous at x = 2.
It is easy to identify points of discontinuity by observing a
graph.
1. If there is point on the graph where the limit does not exist
the graph is
discontinuous.(Remember that the limit only exist if the left
hand limit
and the right hand limit are equal)
2. If the limit exist at x = c, but it is not equal to f(c) the
graph is discontinuous.
3. If f(c) does not exist also the graph is discontinuous.
The following exercise should be useful to strengthen your
understanding of
continuity.
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
14
a. Hole: f(c) is not defined existsxfcx
)(lim
.
b. Hole: f(c) is defined existsxfcx
)(lim
and is not equal to f(c).
c. Jump: )(lim1
xfcx
is not the same as )(lim xfcx
.
d. Pole : f(c) is defined at x = c;
)(lim1
xfcx
e. Pole: f(c) is defined at x = c ;
)(lim1
xfcx
and
)(lim xfcx
Worksheet #1
Find the limits of the following:
1. xx
xx
x 2
124lim
2
2
2
2.
h
h
h
1832lim
2
0
3. t
tt
t
4
43lim
4
4. x
x
x
33lim
0
5. Evaluate
33
24
312
lim3
xx
xx
xfx
Sketch a graph to represent xfx 3lim
6. Using the graph, compute each of the following.
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
15
(i) 4f (ii) 4
lim ( )x
f x
(iii) 1
lim ( )x
f x
7. (a) List the properties of a function which is continuous at
a.
(b) Show that f(x) is continuous at x = -1
State the values of x for which the function is
discontinuous.
(i) f(x) = 103
22
xx
x (ii) g(x) =
16
42
x
x
8. Determine where the function below is not continuous 152
1042
tt
tth
Activity Sheet
Calculus worksheet #2 - Continuity
a. Is 4
16)(
2
x
xxf continuous at x = 4. Give reason(s) for your answer.
b. Find the value of the constant A that makes the given
function f(x) continuous for
all x.
13
11
1
)(2
2
xifxAx
xifx
x
xf
c. Determine if f(x) is continuous at x = -2, where
-
VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2
16
263
24)(
xifx
xifxxf
d. Determine if f(x) is continuous at x = 2, where
2
2
11
5
23
)(4 xif
xif
x
xifx
xf
END OF HANDOUT