AN INTRODUCTION TO CALCULUS THE LIMIT OF A FUNCTION In the previous section, we examined tangents and how to compute their equations. The examples we treated showed that limits arise whenever we attempt to compute velocities of objects, or tangents to curves. In the first example of the previous section, recall that we dealt with the reciprocal function y = 1/x, and tried to compute the equation of the tangent line at point P (0.5, 2). From the tables, we saw that, as x approached 0.5 from either side (that is, from the left and right of 0.5), the values of the slopes of the secant lines approached –4. Thus, we assumed that the slope of the tangent line at P was –4. Consequently, this allowed us to compute the equation of the tangent. From the example and the others, it is clear that the slope of a tangent is the limit of the slope of the secant lines . Assuming we represent the slope of a typical secant line by MPQ, and the slope of the tangent by M, then,symbolically, the situation is expressed like this:
lim MPQ
Q→P
=
M
1
This mathematical representation will be explained in detail as we move on. Similarly, for the velocity problems, we saw that the instantaneous velocity of an object is the limit of average velocities . Here's one way of defining a limit:
Given two quantities x and f(x), we can take x to be sufficiently close to a number a (as close as possible, but not equal to a), so that the values of f(x) will be as close to another number L as possible). And so, we write
lim
x→a
f(x)
=
L
2
which is read as “the limit of f(x) , as x approaches a, is L”.
Thus, equation 1 above can be read as
“the limit of the slope of the secant line MPQ, as Q(x, y) approaches P(0.5, 2), equals the slope of the tangent, M”. In the previous, I mentioned that the idea of one quantity f(x) approaching a value L, as another related quantity x approaches a value a, is the most basic concept of a limit.
Greek scholars like Archimedes and Eudoxus used a “method of exhaustion” to calculate areas and volumes, a method in which limits were indirectly used. Even when calculus was in its developmental stages, Newton's immediate predecessors (like Barrow, Fermat and Cavalieri), did not use limits. Sir Isaac Newton, who was the first mathematician to speak explicitly about limits explained (and I quote): “quantities approach nearer than by any given difference”. He further stated that the limit was the basic concept in calculus, but such ideas were left to later mathematicians like Cauchy to clarify. Study this example carefully.
EXAMPLE 1 Assume you have the function
g(x) =
x–1 x3 – 1
Evaluate g at numbers close to 1 (but NOT EQUAL TO 1), correct to six decimal places. Use the results to guess the value of
x–1
lim
x3 – 1
x→1
Solution The first step to solving this problem is drawing up a table of values.
x
g(x)
Undertand this: when we are dealing with limits, the ONLY thing
0.2
0.806452
that concerns you is how a given function f behaves near a. This
0.4
0.641026
means that f does not nessarily need to be defined at a. In this
0.6
0.510204
case, we're dealing with
0.8
0.409836
0.9
0.369003
0.99
0.336689
lim
x→1
x–1 x3 – 1
Here, we see that a = 1. So, if we put x = 1 in the function above,
Table of values of y = g(x) (as x approaches 1 from the left)
we see that g is undefined, but DOES NOT MATTER, as we are only concerned with how f behaves as x approaches 1 from both sides.
x
g(x)
From the two tables on the left, we might be slightly unsure of
1.8
0.165563
what the limit is. A possible solution is to find the average of the
1.6
0.193799
two closest values of g(x): 0.336689 and 0.330022:
1.4
0.229358
1.2
0.274725
1.1
0.302115
1.01
0.330022
½ (.336689 + 0.330022)
= 0.3333555
Thus, we can say that the limit is about 0.333 (or 1/3). When in doubt, the best alternative is to use even smaller (but NOT TOO SMALL) values of x that approach 1:
Table of values of y = g(x) (as x approaches 1 from the right)
x
g(x)
x
g(x)
0.99
0.336689
1.01
0.330022
0.995
0.335005
1.005
0.331672
0.999
0.333666
1.001
0.333000
0.9995 0.333667
1.001
0.333300
Table of values of y = g(x) (as x approaches 1 from the both sides) On the third table, there is clearly a pattern: as x approaches 1 from both sides, the values of g(x) approach 0.333 (or 1/3). Thus, we say
lim
x→1
x–1 x3 – 1
=
which is read as “ the limit of g(x), as x approaches 1, is 1/3”. Besides the default notation
lim
x→a
f(x)
=
L
an alternative notation is
f(x) → L
as
x→a
1 3
which is quite straightforward: f(x) approaches L as x approaches a. Let's look at another example:
EXAMPLE 2 Evaluate the function
1 – x2
g(x) =
x2 + 3x – 10
at the given numbers (correct to six decimal places): x = 3, 2.1, 2.01, 2.001, 2.0001, and 2.00001. Use the results to guess the value of
lim
x→2
1 – x2 x2 + 3x – 10
or explain why it does not exist.
Solution Like the first example, we draw up a table of values, and observe any pattern that pops up:
x 3
Take a good look at this table, and ask yourself: does a limit exist?
g(x)
Well, you'll notice that the values of g(x) decrease infinitely as x
–1
approaches 2. We therefore express this observation verbally by saying
2.1
–4.802816
2.01
–43.368046
2.001
–429.081703
2.0001
–4286.224497
that the limit of g(x), as x approaches 2, is negative infinity, OR, g(x) decreases infinitely as x approaches 2. Mathematically, we express this limit as
lim g(x)
2.00001 –42857.714290
x→2
=
–∞
This example illustrates a kind of limit, which is called an INFINITE LIMIT. Understand this: the symbol ∞ (the infinity symbol) IS NOT A NUMBER. Thus, the limit
lim g(x)
x→2
=
–∞
does not imply that a limit exists. A limit does not exist in this case. Instead, the equation specifically describes how the
limit does not exist, that is, the values of g(x) can be made as large negative as possible when x gets closer and closer to 2. An infinite limit can either be positive or negative. In other words, an infinite limit can be positive infinity or negative infinity:
lim
x→a
f(x)
lim f(x)
x→a
=
=
∞
3
–∞
4
The table below show a number of ways of expressing infinite limits verbally:
lim
x→a
f(x)
=
lim f(x)
∞
x→a
=
–∞
The limit of f(x), as x approaches a, is infinity
The limit of f(x), as x approaches a, is negative infinity
f(x) increases without bound as x approaches a
f(x) decreases without bound as x approaches a
f(x) becomes infinite as x approaches a
The next example will illustrate another kind of limit: one sided limits. To illustrate one sided limits, we'll use the signum (or sign) function. The sign function (denoted by sgn), is such that
Sgn x
=
–1
if x < 0
0
if x = 0
1
if x > 0
If we graph the sign function, we have this:
From the graph, it is clear that, as x approaches 0 from the left, sgn x approaches –1. Therefore, we say
lim sgn x =
x→0-
–1
5
In equation 5, notice the symbol x→0-. The minus sign is an indication that the values of x approach 0 FROM THE LEFT. When we are computing limits, and we see x→a-, it means we consider values of x less than a Conversely, we see from the graph that sgn x approaches 1 as x approaches 0 from the right.. This limit is mathematically expressed as
lim
x→0+
sgn x =
1
6
In this case, the plus sign in the symbol x→0+ (in equation 6) means that the values of x approach 0 FROM THE RIGHT. In other words, we consider values of x greater than a when computing limits with the symbol x→0+. Thus, the limit
lim f(x) =
L
x→a-
is called a LEFT – HAND LIMIT, and can be interpreted in one of the following ways: ●
the left hand limit of f(x), as x approaches a, equals L, OR
●
the limit of f(x), as x approaches a from the left, equals L.
Similarly, the limit
lim f(x) =
L
x→a+
is called a RIGHT – HAND LIMIT, and can be read as ●
the right hand limit of f(x), as x approaches a, equals L, OR
●
the limit of f(x), as x approaches a from the right, equals L
Note this, If the left hand and right hand limits (of the same function y = f(x)) are not equal, that is, if
lim
f(x)
≠
lim
f(x)
DOES NOT EXIST
x→a-
lim
x→a+
f(x)
then, x→a
Finally, note the the concept of one sided limits also applies to infinite limits. Hence, we have four possible scenarios:
lim f(x) =
–∞
7
lim f(x) =
–∞
8
lim f(x) =
∞
9
lim
∞
10
x→a-
x→a+
x→a-
x→a+
f(x) =
Equations 7 through 10 above introduces us to another concept: the vertical asymptote. By definition, if AT LEAST one of equations 7 through 10 apply to a function y = f(x), then the line x = a is said to be the vertical asymptote to the curve y =f(x). Here's an illustration.
EXAMPLE 4 Find the vertical asymptotes of the function
f(x) =
x x –x–2 2
Solution To find the vertical asymptote(s) of a rational function (such as f above), the best route is to find the values of x that make the function undefined. We know that for a rational function, division by zero is not allowed; in other words, a rational function is undefined when the denominator evaluates to zero. In this case, we need to find the values of x such that x2 – x – 2 = 0. Using the quadratic formula, we see that the denominator equals zero when x = -1, or x = 2. Thus, f is undefined when x = -1, or x = 2. This means that the lines x = -1 and x = 2 are the vertical asymptotes of f. A visual aid can also be of assistance:
Vertical asymptote x = -1 (Red dash-line)
Vertical asymptote x = 2 (Red dash-line)
Note: On graphs, vertical asymptotes are usually represented by vertical dash lines. From the graph, observe the following: ●
f(x) → ∞ as x → –1 from the right.
●
f(x) → – ∞ as x → –1 from the left.
●
f(x) → ∞ as x → 2 from the right.
●
f(x) → – ∞ as x → 2 from the left.
Thus, based on the definition of a vertical asymptote, the lines x = -1 and x = 2 are the vertical asymptotes to the function
f(x) =
x x –x–2 2
EXAMPLE 5 Use a graph of the function
f(x) =
1 1 + 21/x
to state the value of each limit, if it exists. If it does not exist, explain why.
(A)
lim
x→0-
(B)
f(x)
lim x→0+
f(x)
(C)
lim
x→0
f(x)
Solution The question has already told us how to solve the problem: using a graph. So, here it is:
(A)
From the graph, f(x) seems to approach 1 as x approaches 0 from the left. Thus,
lim
x→0-
(B)
This is the left hand limit
On the other hand, f(x) seems to approach 0 as x approaches 0 from the right. This means
lim
x→0+
(C)
f(x) = 1
f(x) = 0
This is the right hand limit
From (A) and (B) above, we see that the left hand and right hand limits are not equal. Thus
lim
x→0
f(x)
DOES NOT EXIST
EXAMPLE 6 Sketch the graph of the following function and use it to determine the values of a for which
lim
x→a
f(x)
exists.
f(x)
=
2–x
if x < –1
x
if –1 ≤ x < 1
(x – 1)2
if x ≥ 1
Solution We are asked to evaluate the limit of a piecewise defined function. The first step should be to graph the function so as to get a visual of what we are dealing with:
Before we start, let's identify the three lines, as defined by the function f: ●
The blue curve represents y = 2 – x
●
The red curve represents y = x.
●
The yellow curve represents y = (x – 1)2
Notice that the curve “breaks” at x = -1 and x = 1. These breaks are called discontinuities. Generally, a limit does not exist at a point where there is a discontinuity. We will examine continuity of functions in the next section. Let's go back to the graph. From the graph, we see that
lim f(x)
x→–1 –
= 3
AND
lim f(x)
x→–1 +
= –1
So, since the left and right hand limits are not equal, it means
lim
X→ –1
f(x)
DOES NOT EXIST
Also, we see that
lim f(x)
x→1 –
= 1
lim f(x)
AND
x→1 +
= 0
Again, the left and right hand limits are not equal, which means
lim
f(x)
X→1
DOES NOT EXIST
Hence, for the function f, the two limits that DON'T EXIST are
lim
X→ –1
f(x)
lim
AND
X→1
f(x)
In other words, the limit
lim
x→a
f(x)
exists for ALL values of a in the domain of f except –1 and 1. Therefore, the values of a for which the limit exists is
{x|x ≠ –1, x ≠1}
EXAMPLE 7 Evaluate the function
(1/√x) – 0.2 x – 25
F(x) =
at the given numbers: x = 26, 25.5, 25.1, 25.05, 25.01, 24, 24.5, 24.9, 24.95, 24.99. Use the result to guess the value of the limit, or explain why it does not exist.
lim
x→25
(1/√x) – 0.2 x – 25
Solution As usual, we use the given values of x to construct a table of values:
x
F(x)
x
F(x)
26
– 0.00388386
24
– 0.00412415
25.5
– 0.00294098
24.5
– 0.00406102
25.1
– 0.00398880
24.9
– 0.00401204
25.05 – 0.00399401
24.95
– 0.00400601
25.01 – 0.00399800
24.99
– 0.00400120
From the table of values, it appears that the values of F(x) approach –0.004 as x approaches 25 from either side. Thus, we say that
lim
x→25
(1/√x) – 0.2 x – 25
=
–0.004
You can use values of x closer to 25 to confirm your assumptions. You can also make use of a graph.
EXAMPLE 8 Evaluate the function
1 – cos x
f(x) =
x2
at the given numbers: x = 1, 0.5, 0.4, 0.3, 0.2, 0.1, 0.05, 0.01. Use the result to guess the value of the limit, or explain why it does not exist.
1 – cos x
lim
x2
x→0
Solution Using the given values of x, we have the following table of values:
x
f(x)
1.0
0.459698
0.5
0.489669
0.4
0.493368
0.3
0.496261
0.2
0.498336
0.1
0.499583
0.05
0.499896
0,01
0.499996
From the table of values, it is evident that f(x) approaches 0.5 as x approaches 0. In fact, if you put x = 0.0001, f(x) is exactly 0.5. Therefore, we say that
lim
x→0
1 – cos x
=
x2
0.5
EXAMPLE 9 Determine the value of the limit.
6
lim
x→5
x–5
Solution To determine the value of the limit, we draw up a table of values using values of x that approach 5 FROM BOTH SIDES. It is important that we compute left and right hand limits, because, as I mentioned earlier, a limit exists if and only if the left and right hand limits are equal. With that in mind, we have the following table:
x
f(x)
x
f(x)
6
6
4
–6
5.5
12
4.01
–6.06
5.1
60
4.05
–6.32
5.01
600
4.5
–12
5.001
6000
4.55
–13.33
5.0005
12000
4.9
–60
5.0001
60000
4.995
–1200
5.00001
600000
4.999
–6000
5.000001
6000000 4.9995
–12000
The first column shows values of x that approach 5 from the right. We see that the corresponding values of f(x) get larger and larger. In other words, the values of f(x) increase infinitely as x approaches 5 from the right. Thus, we have the right hand limit
lim
6
x→5+
x–5
=
∞
On the other hand, we have the third column which shows values of x that approach 5 from the left. Notice that the corresponding values of f(x) decrease infinitely. Therefore, we have the left hand limit.
lim
6
x→5+
x–5
=
–
∞
So, the left hand limit is negative infinity, and the right hand limit is positive infinity. The graph also confirms our assumptions:
Notice that the curve “breaks” at x = 5. Since
6
lim x→5+
=
x–5
∞
AND
lim
6
x→5+
x–5
it means
lim
x→5
6
DOES NOT EXIST
x–5
It also means that the line x = 5 is a vertical asymptote to the curve
6 x–5
f(x) =
(as indicated by the blue vertical dash line in the graph above).
Determine the values of a for which the limit
x –1
lim
x→a is infinite.
x (x + 2) 2
EXAMPLE 10
=
–
∞
Solution We can start by graphing the function
f(x) =
x –1 x (x + 2) 2
If an infinite limit exists, we should expect at least one discontinuity, and indeed, there is/are:
From the graph, there is a discontinuity at two values of x: –2 and 0. So, let's work with these two values of x and see if a limit exists at both points. The table below shows the values of f(x) for values of x close to – 2:
x
f(x)
x
f(x)
–1
–2
–2.5
1.12
–1.1
–1.9284
–2.1
7.0295
–1.2
–1.9097
–2.05
14.5152
–1.5
–2.2222
–2.01
74.5031
–1.8
–4.3209
–2.005
149.5016
–1.95
–15.5161
–2.001
749.5003
–1.99
–75.5031
–2.0005
1499.5002
–1.9995
–1500.5001
–2.0001
7499.5000
–1.9999
–7500.5
–2.00001
74999.5
–1.99995
–15000.5
–2.000001
749999.5
–1.99999
–75000.5
–2.0000001 7499999.5
From the table, it is evident that f(x) decreases infinitely as x approaches –2 from the right. Also, we see that f(x) increases without bound as x approaches –2 from the left. Thus, we have two infinite limits:
x –1
lim
x→–2 –
x (x + 2)
lim
x –1
2
=
∞
AND x→–2+ x2(x + 2)
=
–
∞
The results above are an indication that
x –1
lim
x→2
x (x + 2) 2
does not exist. Instead, we have two infinite limits at x =2:
lim
x→–2
x –1 –
∞
=
x (x + 2) 2
x –1
lim
AND
=
x→–2+ x (x + 2) 2
–
∞
It also indicates that the line x = 2 is a vertical asymptote to f. Next, we try values of x close to 0 and see how f responds:
x
f(x)
x
From the table, f(x) decreases without bound as x approaches 0from BOTH SIDES. Thus, we have the infinite limits
f(x)
–1
–2
1
0
–0.9
–2.1324
0.9
–0.04257
–0.5
–4
0.5
–0.8
–0.3
–8.4967
0.3
–3.3816
–0.1
–57.8947
0.1
–42.8571
–0.015
–2272.6
0.015
–2172.5944
–0.001
–500750.3752
0.001
–4925.3731
–0.0001
–50007500.375
0.0001
–49992500.375
–0.00001
–4999925000.3
0.00001
–499925000.3
0.0.000001
–4.999 × 10
–0.0.000001 –4.999 × 10
11
lim
x –1
x→0+ x2(x + 2)
=
–
∞
=
–
∞
AND
11
lim
X→0−
x –1 x2(x + 2)
This indicates that the line x = 0 (which is the y-axis itself) is the second vertical asymptote to the function f.
So, we have four infinite limits
x –1
lim
x→–2 –
x –1
lim
x→–2+ x (x + 2) 2
∞
=
x (x + 2) 2
=
–
x –1
lim
x→0+ x2(x + 2)
x –1
lim
∞
X→0−
x (x + 2) 2
=
–
∞
=
–
∞
These infinite limits occur when x = –2 or x = 0. And as the graph clearly shows, the lines x = –2 and x = 0 (the y-axis) are vertical asymptotes to the function
f(x) =
x –1 x (x + 2) 2
EXAMPLE 11 Determine the infinite limit.
lim csc x
x→π–
Solution We know that
lim csc x
=
x→π–
lim
x→π–
1 sin x
Therefore, we have the following table:
θ
x (Radians)
From the table, it is clear that f(x) does not approach any particular number as x approaches π from the left. In other words, f(x) increases infinitely as x gets closer and closer to π (from the left). This means
f(x)
0
0
∞
30
π/6
2
45
π/4
1.414213
60
π/3
1.154700
90
π/2
1
120
2π/3
1.154700
150
5π/6
2
160
8π/9
2.923804
170
17π/18
5.758771
175
35π/36
11.473713
176
44π/45
14.335587
179
179π/180
57.298689
179.9
1799π/1800
572.958086
179.95
3599π/3600
1145.915736
lim csc x
lim
1 sin x
x→π–
=
∞
The table expresses the angles in degrees, to make the calculations easier. To convert between Radians and degrees, remember that
10
=
1 rad
=
π
180
rad
180 0
π
EXAMPLE 12
Determine the infinite limit.
x→(–π/2)–
lim
=
x→π–
sec x
Solution Since
θ
lim
x→(–π/2)–
sec x
=
lim
1 cos x
x→(–π/2)–
We put
g(x) =
1 cos x
and generate the following table of values. Understand that we are looking for the left hand limit only. The table shows that the values of g(x) decrease without bound as x approaches –π/2 (which corresponds to –90 0) from the left. Hence, the infinite limit is
lim
x→(–π/2)–
sec x
=
–
∞
x (Radians)
g(x)
–180
–π
1
–170
–17π/18
–1.015427
–160
–8π/9
–1.064178
–150
–5π/6
–1.154700
–120
–2π/3
–2
–100
–5π/9
–5.758771
–90.9
–101π/200
–63.664595
–90.01
–9001π/18000
–5729.577981
–90.001
–90001π/180000
–57295.779548
–90.0001
–9000001π/1800000
–572957.79578
EXAMPLE 13
Determine the infinite limit.
lim
x+1 x sin πx
x→1+
Solution As usual, we draw up a table of values. Since we are to compute the right hand limit only, we use values of x that approach 1from the right.
x
f(x) Notice that f(x) decreases infinitely as x approaches 1 from the right. Thus, the limit is negative infinity. In other words,
1.5
–1.666666
1.1
–6.177945
1.05
–12.480504
lim
1.01
–63.357240
x→1+
1.001
–636.302827
1.0001
–6365.879551
1.00001
–63661.658952
1.0000001
–6366197.5139
lim
=
–
∞
EXAMPLE 14
Determine
X→1–
x+1 x sin πx
1 x3 – 1
lim
and
X→1+
1 x3 – 1
(a) by evaluating f(x) = 1/(x3 – 1) for values of x that approach 1 from the left and from the right, (b) by reasoning, and (c) from a graph of f.
Solution (a) Take a good look at the question, and you'll see that we are being asked to evaluate the limits using three different methods. In the first instance, we want to evaluate the limits using values x that approach 1 from the left and from the right, This is the method that involves the use of a table of values.
x
f(x)
x
f(x)
0.5
–1.142857
1.5
0.421052
0.8
–2.049180
1.4
0.573394
0.9
–3.690037
1.1
3.021148
0.99
–33.6689
1.01
33.002211
0.995
–67.001114
1.001
333.000222
0.999
–333.666889
1.0001
3333.000022
0.9995
–667.000111
1.00001
33333.000003
0.9999
–3333.666689
1.000001
333333
NOTE: On the table, the first two columns show values of x close to, but less than 1 and the corresponding values of f(x) While the third and fourth columns show values of x close to, but greater than 1 and the corresponding values of f(x)
From the table, we see that, as x gets closer to 1 from the left, the values of f(x) decrease infinitely. This means the left hand limit is negative infinity:
lim X→1–
1 x3 – 1
=
–
∞
On the other hand, as for the right hand limit, we see that the values of f(x) increase without bound as x approaches 1 from the right. Thus, the value of the right hand limit equals
lim
1 x3 – 1
X→1+
=
∞
These infinite limits mean that the line x = 1 is a vertical asymptote to f.
(b) The second part of the question asks us to evaluate the limits by reasoning. Take a good look at this limit:
lim X→1–
1 x3 – 1
This is a left hand limit, which would require that we evaluate it using values close to, but less than 1. Our first focus should be the denominator, since that is where x is. Suppose x is close to 1, but less than 1. This would mean that the denominator would be a very small negative number, and thus f would evaluate to a large negative number. In other words, f becomes numerically larger (negatively) as x gets closer and closer to 1. Intuitively this would imply that the limit is negative infinity (–∞). A similar scenario occurs when we want to evaluate the right hand limit
lim X→1+
1 x3 – 1
Here, we use values close to, but greater than 1 this time around. So, assume that x is greater than 1. this would make the denominator a numerically small positive number, which, in turn, makes f a large positive number. Thus, as long as x gets closer and closer to, but remains greater than 1, then f gets bigger and bigger (positively). Therefore we have a positive infinite limit.
(c) The third part of the question asks that we evaluate the limits by graphing the function f:
Vertical asymptote x = 1 (blue dash-line)
Take a good look at the curve, and observe the part of the curve that lies between the positive x and y axes. You can see that the curve extends upwards indefinitely, but NEVER touches the line x = 1 (the vertical dash line). Thus, as x approaches 1 from the right, the values of f(x) increases without bound. This is therefore a positive infinite limit, and so,
lim
1 x3 – 1
X→1+
=
∞
On the other side of the vertical dash line, observe that the curve extends downwards infinitely, and does not touch the line x = 1. Therefore, this would mean that the values of f(x) decrease without bound as x approaches 1 from the left. Thus we have a negative infinite limit:
lim X→1–
(a)
1 x3 – 1
=
–
∞
EXAMPLE 15
Estimate the value of the limit
lim (1 + x)1/x
x→0 (b)
Illustrate part (a) by graphing the function y = (1 + x)1/x.
Solution To estimate the limit, we draw up a table of values, use the table to deduce the left and right hand limits. From there, we'll be able to derive the value of the limit:
x
f(x)
x
f(x)
–0.9
12.915497
1
2
–0.5
4
0.5
2.25
–0.15
2.954884
0.15
2.538939
–0.1
2.867972
0.1
2.593742
–0.01
2.731999
0.01
2.704813
–0.001
2.719642
0.001
2.716924
–0.0001
2.718146
0.0001
2.718146
–0.00001
2.718268
0.00001
2.718268
–0.000001
2.718283
0.000001
2.718280
0.0000001
2.718282
–0.0000001 2.718282 From the table, it is evident that
lim (1 + x)1/x
≈
2.718
(left hand limit)
lim (1 + x)1/x
≈
2.718
(right hand limit)
=
2.718
X→0– and
X→0+
Therefore, we can conclude that
lim (1 + x)1/x
X→0–
Ordinarily, we find that the function y = (1 + x)1/x is undefined when we put x = 0, but don't let that throw you off track, as we are only concerned with how this function behaves as it approaches 0. The function is graphed below:
Using the trace function on a graphing calculator or computer, we see that f(x) approaches 2.718 (approx) from either side of the y-axis. Hence, from the table above and the graph, we assume that
lim (1 + x)1/x
X→0–
=
2.718
EXAMPLE 16 The slope of the tangent line to the graph of the exponential function y = 2x at the point (0, 1) is
lim (2x – 1)/x
X→0 Estimate the slope to three decimal places.
Solution From the question, the slope of the tangent line is expressed explicitly by a formula
lim (2x – 1)/x
X→0
which happens to be a limit in this case. Since the task is to estimate the slope, it means we evaluate the given limit. We start by constructing a table of values for g(x) = 2 x – 1 for several values of x that approach 0. (The table is on the next page). The table clearly displays a pattern: the values of g(x) approach 0.693 as x gets closer and closer to zero (from both sides). This indicates the equality of both left and right hand limits, and so, we guess that
lim (2x – 1)/x
X→0
≈
0.693
x
g(x)
x
g(x)
1
1
–1
0.5
0.5
0.828427
–0.5
0.585786
0.4
0.798769
–0.4
0.605354
0.3
0.770481
–0.3
0.625825
Recall that the limit
lim (2x – 1)/x
X→0
0.1
0.717734
–0.1
0.669670
0.05
0.705298
–0.05
0.681273
represents the slope of the tangent line to the curve y = 2 x.
0.01
0.695555
–0.01
0.690750
0.001
0.693387
–0.001
0.692907
Since this limit evaluates to approximately 0.693, it means that the slope of the tangent line is about 0.693.
0.0001
0.693171
–0.0001
0.693123
0.00001
0.693149
–0.00001
0.693144
0.000001
0.69315
–0.000001
0.693147
0.0000001
0.6931
–0.0000001 0.69314
EXAMPLE 17 (a)
By graphing the function f(x) = (tan4x)/x and zooming in toward the point where the graph crosses the y-axis, estimate the value of
lim (tan 4x)/x
X→0 (b)
Check your answer in part (a) by evaluating f(x) for values of x that approach 0.
Solution If we graph the function in the viewing rectangle [–3, 3] by [–10, 10], we get this:
On the graph above, notice the blue box, this is where the curve intersects the y-axis. If we zoom in toward this area (we zoom in till we reach the viewing rectangle [–0.3, 0.3] by [–10, 10]) , we have this:
f(x)→4 as x →0
The graph is clear enough in this viewing rectangle. From the graph, it is very obvious that f(x) approaches 4 as x approaches 0 from both sides. Therefore, we say that
lim (tan 4x)/x
X→0
=
4
So, using a graph, we were able to deduce the value of the limit is 4. Now, let's verify this guess using numerical evidence. In other words, we evaluate f(x) for values of x that approach zero from both sides. Thus, we have the following table
x
f(x)
x
f(x)
1
1.157821
–1
1.157821
0.5
–4.370079
–0.5
–4.370079
0.4
–85.581331
–0.4
–85.581331
0.3
8.573838
–0.3
8.573838
0.2
5.148192
–0.2
5.148192
0.1
4.227932
–0.1
4.227932
0.05
4.054200
–0.05
4.054200
0.01
4.002134
–0.01
4.002134
0.001
4.000021
–0.001
4.000021
0.0001
4.000000
–0.0001
4.000000
0.00001
4.000000
–0.00001 4.000000
The values on the table clearly speak for themselves: as x approaches zero from both sides, the values of f(x) also get closer and closer to 4. In other words
lim (tan 4x)/x
=
4
lim (tan 4x)/x
=
4
lim (tan 4x)/x
=
4
X→0– and
X→0+ which means X→0
If we put x = 0, you'll see that f is undefined, because the denominator would be zero, but it does not matter, since we are only concerned with how f behaves as it approaches zero.
NOTE: If you use a graphing calculator or computer graphing software, then you should be familiar with the concept of the
viewing rectangle. The viewing rectangle is simply the rectangular viewing screen/display window where a graph is displayed. A viewing rectangle is defined by
[a, b]
×
[c, d]
where
(a)
a =
minimum value of x-axis
b =
maximum value of x-axis
c =
minimum value of y-axis
d =
maximum value of y-axis
Estimate the value of
lim X→0
EXAMPLE 18
6x – 2x x
by graphing the function g(x) = (6x – 2x)/x. (b)
Check your answer in part (a) by evaluating f(x) for values of x that approach 0.
Solution The first part asks that we evaluate the limit by making use of a graph. Understand that g is undefined when x = 0, but when we graph the function, we get
Using the trace function on a graphing calculator or computer, we find that, as x gets closer and closer to zero, the values of g(x) approach 1.099 (approx).
Using a table of values, we can make a more accurate guess:
x
g(x)
x
g(x)
The table clearly shows that
0.5
2.07055
–0.5
0.59772
0.4
1.82041
–0.4
0.67375
0.3
1.60208
–0.3
0.76021
0.2
1.41135
–0.2
0.85862
0.1
1.24458
–0.1
0.97074
lim X→0+
0.01
1.11235
–0.01
1.08505
0.001
1.09998
–0.001
1.09725
0.0001
1.09875
–0.0001
1.09848
0.00001
1.09863
–0.00001
1.09859
0.000001
1.09861
–0.000001
1.09861
0.0000001
1.0987
–0.0000001
1.09861
0.00000001 1.099
lim X→0–
6x – 2x x
≈
1.099
6x – 2x x
≈
1.099
and
Therefore, we conclude that
lim X→0
6x – 2x x
=
1.099
–0.00000001 1.0986
EXAMPLE 19
Evaluate the function f(x) = x2 – (2x/1000) for x = 1, 0.8, 0.6, 0.4, 0.2, 0.1, and 0.05, and guess the value of
(a)
lim X→0
(b)
x2
–
2x 1000
Evaluate f(x) for x = 0.04, 0.02, 0.01, 0.005, 0.003, and 0.001. Guess again.
Solution Using the given values of x, we have the following table of values:
x
f(x)
From the table, it appears that, as x approaches 0 from the right, the values of f(x) seem to get closer to zero.
1
0.998
0.8
0.638258
0.6
0.358484
0.4
0.158680
lim
0.2
0.038851
X→0
0.1
0.008928
0.05
0.001465
If you are not so sure, you can average the two closest values of f(x). Either way, we find that
–
x2
2x 1000
=
0
The next question asks that we evaluate f(x) for values of x that are considerably closer to zero than the first set of numbers. Hence, we have this table:
x
f(x)
0.04
0.000571
0.02
0.000613
0.01
0.000907
0.005
0.000978
0.003
0.000993
0.001
0.000999
What we have on this table is considerably different from the results on the previous table. In this case, we see that as x gets closer and closer to 0, the values of f(x) also get closer and closer to 0.001 (approx). Therefore,
lim X→0
x2
–
2x 1000
=
0.001
Below is a graph the function f(x) = x2 – (2x/1000):
If you are familiar with quadratic equations and graphs, then you'll agree the graph above looks pretty much like the graph of y = x2. Clearly, the graph shows that the limit is indeed zero. So, why did we get 0.001 in the second calculation? When we are trying to guess limits using computers and calculators, there are two potential problems we face: ●
If we use inappropriate values of x, chances are that we will guess the wrong value of a limit, and
●
Sometimes, it can be difficult to determine when to stop calculating values.
So, we got 0.001 as the value of the limit in the second table perhaps because we used inappropriate values of x.
This example has illustrated the huge drawbacks in guessing the values of limits. In the following section, we will use standard, fail-safe methods, known as LIMIT LAWS to evaluate limits.
EXAMPLE 20
(a)
Evaluate h(x) = (tan x – x )/x3 for x = 1, 0.5, 0.1, 0.05, 0.01, and 0.005.
(b)
Guess the value of
lim X→0
(c)
tan x – x x3
Evaluate h(x) for successively smaller values of x until you finally reach 0 values for h(x). Are you still confident that your guess in part (b) is correct? Explain why you eventually obtained 0 values.
(d)
Graph the function h in the viewing rectangle [–1, 1] by [0, 1]. Then zoom in toward the point where the graph crosses the y-axis to estimate the limit of h(x) as x approaches zero. Continue to zoom in until you observe distortions in the graph of h. Compare with the results of part (c).
Solution We evaluate h for the given values of x:
x
h(x)
1
0.557408
0.5
0.370419
0.1
0.334672
0.05
0.333667
0.01
0.333346
0.005
0.333337
From the table, it appears that h(x) approaches 1/3 as x approaches 0. Therefore,
lim X→0
tan x – x x3
=
1 3
Next, we evaluate h even smaller values of x until we eventually reach 0 values for h(x):
x
h(x)
It is very clear that something strange is happening. The limit
0.001
0.333333
lim
0.00001
0.3333
X→0
0.000005
0.33328
0.000001
0.33
0.0000005
0.328
0.0000001
0
tan x – x x3
does exist (as we will see later). Again, we have fallen prey to the calculator problem. The limit is really 1/3. But we have obtained zero values because both numerator and denominator approach 0 as x gets closer and closer to zero. And as you can see from the table, the values of x are pretty small.
In other words, let's assume that f(x) = tan x – x g(x) = x3 So that h(x) = f(x)/g(x). Because the values of x are very small, f(x) → 0 and g(x) → 0 as x → 0. Thus, this limit would equal zero.
Therefore,
this error can be attributed to the values of x that were used; they were too small, and
ultimately led to problems with subtraction when using a calculator to calculate values. If we graph h in the viewing rectangle [–1, 1] by [0, 1], we have this:
Observe the blue box; this is where the crosses the y-axis. If we zoom in toward this area, this Is what we get:
The graph above shows h(x) = (tan x – x )/x3 in the viewing rectangle [–0.01, 0.01] by [ –0.05, 1.04]. Using the trace function on a graphing calculator or computer, we see that
lim X→0
When we zoom in too far, we get this:
tan x – x x3
=
1 3
Here, the function is graphed in the viewing rectangle [–0.00000001, 0.00000001] by [ –0.05, 1.04]. The graph becomes inaccurate because of problems associated with subtraction. It gets worse when we zoom in even further:
This time, h is graphed in the viewing rectangle [–0.000000001, 0.000000001] by [–0.05, 1.04]. Using this graph, it would seem that the limit is really zero, but it's not because, the values of x are too small. Thus, when the graphing calculator/computer calculates the values, they would be read as zero. Here's one way of looking at the situation: consider a number like 0.000000000001. This number is practically zero isn't it? Well, it is, and that's exactly how a computer would treat it; as if it were zero.
EXAMPLE 21 In the theory of relativity, the mass of a particle with velocity v is
m
=
mo
√1
– v2/c2
where mo is the rest mass of the particle at rest and c is the speed of light. What happens as v→c– ?
Solution We will have to solve this problem using by reasoning, like we did in example 14. The function we're dealing with is
m
=
mo
√1
– v2/c2
The question tells us that c represents the speed of light, which is approximately equal to 300000 km/s. We also assume that the rest mass of the particle is constant. Hence, the velocity v is the only variable in the function.
The main focus here should be the denominator:
√1
– v2/c2
The question is asking us to determine what happens as v→c– . In other words, we want to know what happens as the speed of the particle approaches the velocity of light. This can be mathematically interpreted as
mo
lim v→c–
√1
– v2/c2
From the denominator, it is clear that if v is close to, but less than c, then v2/c2 would approach 1. Invariably, the denominator approaches zero, which means the limit DOES NOT EXIST. Let's break it down: As
v → c–,
then
v2/c2 → 1
which means
√1
– v2/c2
→
0
and so,
mo
lim v→c–
√1
=
– v2/c2
∞
EXAMPLE 22
Use a graph to estimate the equations of all the vertical asymptotes of the curve
y = tan(2 sin x)
–π ≤ x ≤ π
Solution The function y = tan(2 sin x) has several asymptotes (see the graph below):
(THE RED DASH LINES REPRESENT THE VERTICAL ASYMPTOTES). Within the domain – π ≤ x ≤ π , the function has FOUR vertical asymptotes:
Using a calculator, estimates of the values of these four asymptotes are x = – 2.2439 x = – 0.8975 x = –2.2439 x = 0.8975 Approximately, we have
x = ± 2.2439
and
x = ± 0.8975
EXAMPLE 23 Graph the function f(x) = sin (π/x) in the viewing rectangle [–1, 1] by [–1, 1]. Then zoom in toward the origin several times. Comment on the behavior of this function.
Solution Figure 1 (on the following page) shows what we get when we graph f(x) = sin (π/x) in the viewing rectangle [–1, 1] by [–1, 1]. Figure 2 shows f when we zoom in toward the origin (0,0). Here, you'll see that the curve begins to straighten. If we keep zooming in, we see that the curve is practically vertical. Note that the vertical lines do not touch the origin. This indicates infinite oscillations as x gets closer and closer to zero from both sides of the y-axis.
FIGURE 1: Graph of f in the viewing rectangle [–1, 1] by [–1, 1] . Notice the infinite oscillations around the origin.
FIGURE 2: Graph of f in the viewing rectangle [–0.0000000001, 0.0000000001] by [–2, 2]. Notice the infinite oscillations around the origin.; they increase steadily as x approaches zero.
EXAMPLE 24
Sketch the graph of an example of a function f that satisfies ALL of the given conditions:
(A)
lim f(x)
X→3+
(D)
f(3)
=
= 3
4
(B)
(E)
lim f(x)
=
X→3–
f(–2) =
2
(C)
lim f(x)
X→–2 –
=
2
1
Solution There are several possible answers for this kind of problem. The key is that the graph satisfies all the given conditions. Using our knowledge of limits and how they are graphically interpreted, here's one graph:
f(x) → 4 as x → 3 from the right
f(x) → 2 as x → 3 from the left
f(–2) = 1
f(3) = 3
The orange colored lines represent the function f. The graph has been generated based on the five conditions given in the question. Here's another example:
This graph is similar to the first one, in that it also satisfies the five conditions given in the question. The only difference is that we make use of curves in this graph, unlike the first where only straight lines were used.
EXAMPLE 25 Sketch the graph of an example of a function f that satisfies ALL of the given conditions:
(A)
lim f(x)
=
1
(B)
lim f(x)
=
1
(E)
X→0–
(D)
X→2+
lim f(x) =
–1
X→0+
f(2)
=
1
(C) (E)
Solution
lim f(x)
X→2–
f(0)
=
0
is undefined
Like the previous example, we draw a graph of f based on the six conditions given above. An example is graphed above. Let's apply the limit concept to a real life situation:
EXAMPLE 26 A patient receives a 150-mg injection of a drug every 4 hours. The graph shows the amount of the drug f(t) of the drug in the bloodstream after t hours. Find these limits.
lim f(t)
t→12–
and
lim f(t)
t→12+
Solution To find the values of the limits, we need to understand the graph. The graph is a visual representation of what is going on in the patient's bloodstream. Here's a verbal description: The patient receives the first 150-mg injection, and the concentration of the drug in the bloodstream becomes 150mg. In time, the body will utilize the drug, which means that the amount of the drug will drop over time. Understand that the graph represents the the amount of the drug after t hours. Thus, the graph shows that, after 4 hours, the amount of the drug in the patient's body has dropped to about 70 to 80 mg. Then there's the second dose: another 150 mg. This second dose, plus the remaining amount of the drug from the first injection, brings the total amount to about 200 to 230 mg. After another 4 hours, it drops to about 130 mg. Now, the patient is given the third dose: another 150 mg. Combined with the amount from the second injection, this brings the total amount of the drug to about 280 – 290 mg. Again, it drops, to exactly 150 mg after 4 hours. This means that after 12 hours, the amount of the drug in the patient's body is 150 mg, which was the original amount of the drug administered. Given the fourth dose, the amount of the drug jumps to 300 mg. After another four hours, it drops to about 160 - 180 mg.
Thus, if we read (or trace) the curve from the left, we see that
lim f(t)
t→12–
=
150
Similarly, if we read the curve from the right, we see that
lim f(t)
t→12–
=
300
Before we conclude this section, try the following problems:
PROBLEM 1: Evaluate the function at the given numbers. Use the results to guess the value of the limit, or explain why it does not exist. 3
F(t) =
(A)
√t – 1
√t – 1
(B)
x = 1.5, 1.2, 1.1, 1.01, 1.001.
lim
3
t→1
√t – 1
√t – 1
PROBLEM 2: Determine the infinite limit:
1
lim
x→3
(x – 3)8
PROBLEM 3: Use numerical and graphical evidence to guess the value of the limit
lim
x→1
x3 – 1 √x – 1
calculus4engineeringstudents.com
g(x) =
cos x – 1 sin x
x = 1, 0.5, 0.4, 0.3, 0.2, 0.1, 0.05, 0.01.
lim cos x – 1
x→0
sin x