Chapter 4.5 – Tutorial – Limits at Infinity
In the Chapter 3.5 tutorial, we will cover the topics listed below. If you need further
explanation, let me know and I will expand the tutorial.
•
Determine (finite) limits at infinity.
•
Determine the horizontal asymptotes.
•
Determine infinity limits at infinity.
Example 1:
Solution: lim
x →∞
Find lim
x →∞
6x − 5
2 x 2 + 11
6x − 5
2 x 2 + 11
=
.
∞
is undefined (has no defined meaning). Let’s
∞
1
manipulate the function by multiply it by “1” in the form
x2 .
1
x2
1
lim
x →∞
6x − 5
6x
2
⋅
2
2 x + 11
x = lim
x →∞
1
x2
5
x
−
5
x →∞
x →∞ x
= lim
=
2
x →∞
11
2 x 11
lim 2 + lim 2
+ 2
2
x →∞
x →∞ x
x
x
6−
lim 6 − lim
=
5
5
x
x = lim
x
2
2
x →∞
2 x + 11
2 x + 11
2
x2
x
2
6−0
6
=
2+0
2
2
6−
Let’s verify our answer by evaluating the function at very large numbers.
f (15000 ) =
f (100000 ) =
6 (15000 ) − 5
2 (15000 )2 + 11
6 (100000 ) − 5
2
2 (100000 ) + 11
Note
=
89995
≈ 4.2424
45000011
=
599995
≈ 4.2426
20000000011
6
≈ 4.2426
2
Example 2:
Find lim sin x .
x →∞
Solution: Since the sine function ranges from -1 to 1 every 2π units, it will never
converge to a single real number. This means that the limit does not
exist.
Example 3:
sin x
.
x →∞
x
Find lim
Solution: Since the sine function in the numerator ranges from -1 to 1 every 2π
units and the denominator will approach ∞ , it appears that the fraction
(function) will approach zero. Let’s now try to verify this analytically. We
know that −1 < sin x < 1 .
Divide all three parts of the inequality by x (we don’t have to worry about x being
negative (which would change the direction of the inequalities) because we are
heading toward + infinity.
−
1 sin x 1
<
<
x
x
x
Let’s now take the limit of all three parts of the inequality as x → ∞ .
 1
 sin x 
1
lim  −  < lim 
 < lim
 
x →∞ 
x
→∞
x
→∞
x
 x 
x
 sin x 
0 < lim 
<0
x →∞ 
x 
sin x 
This forces lim 
 = 0.
x →∞ 
x 
For example, f (100, 000 ) =
Example 3:
sin100, 000 0.0357488
=
≈ 0.000000357 ≈ 0
100, 000
100, 000
Suppose that f ( t ) measures the oxygen level in a fish pond, where
t is measured in weeks and f ( t ) = 1 represents the normal unpolluted
level. When t = 0 , organic waste is dumped into the fish pond, and as
the waste material oxidizes, the oxygen level in the pond is
represented by f ( t ) =
t 3 − 2t 2 + 3t + 1
.
t3 + t + 5
Suppose the organic waste
was dumped into the pond on January 1, what is the oxygen level after
1 week? 5 week? 10 weeks? 100 weeks? 1000 weeks? From this
data, what can you conclude about the oxygen level as the number of
weeks approaches infinity (assuming nothing else is dumped into the
pond and the pond does not dry up). Write a paragraph about the
status of the pond after the organic waste is dumped into the pond.
How many weeks will it take before the oxygen level in the pond is
within 5% of the normal level?
Solution: Finding out the oxygen level at weeks 1, 5, 10, 100, and 1000 is just a
matter of substituting these values in the oxygen equation.
•
( )3 − 2 (1)2 + 3 (1) + 1 3
f (1) = 1
= ≈ 0.42857 ≥ 42.9%
7
(1)3 + (1) + 5
•
( 5 )3 − 2 ( 5 )2 + 3 ( 5 ) + 1 91
f ( 5) =
=
≈ 0.67407 ≥ 67.4%
135
( 5 )3 + ( 5 ) + 5
•
(10 )3 − 2 (10 )2 + 3 (10 ) + 1 831
f (10 ) =
=
≈ 0.8187 ≥ 81.9%
1015
(10 )3 + (10 ) + 5
•
(100 )3 − 2 (100 )2 + 3 (100 ) + 1 980301
f (100 ) =
=
≈ 0.980198 ≥ 98.2%
1000105
(100 )3 + (100 ) + 5
•
(1000 )3 − 2 (1000 )2 + 3 (1000 ) + 1 998003001
f (1000 ) =
=
≈ 0.9980 ≥ 99.8%
100001005
(1000 )3 + (1000 ) + 5
It appears that as time goes on, the oxygen level of the pond will return to normal.
To find out how many weeks it will take for the oxygen level to return to within 5% of
normal, we find the real number x such that
t 3 − 2t 2 + 3t + 1
= 0.95 . It appears from
t3 + t + 5
the above substitutions that it will take between 10 and 100 weeks. To be more
specific,
( )3 − 2 ( 50 ) 2 + 3 ( 50 ) + 1 120151
f ( 50 ) = 50
=
≈ 0.96079 ≥ 96.1% This narrows the range
( 50 )3 + ( 50 ) + 5
125055
from 10 to 50.
( 40 )3 − 2 ( 40 )2 + 3 ( 40 ) + 1 60921
f ( 40 ) =
=
≈ 0.95122 ≥ 95.1% This narrows the range
( 40 )3 + ( 40 ) + 5
64045
from 10 to 40 and very close to 40. As we continue this process, we see that it will
take about 39 weeks for the oxygen level to return to 95%.
f ( 39 ) =
( 39 )3 − 2 ( 39 )2 + 3 ( 39 ) + 1 56395
=
≈ 0.95000253 ≥ 95%
59363
( 39 )3 + ( 39 ) + 5
How did we come up with the number 39? We could have kept on substituting until
we found it or we could have graphed
t 3 − 2t 2 + 3t + 1
− 0.95 and look for the tt3 + t + 5
intercepts. As you can see, there is one t-intercept and it is 39.
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