Download Limits at Infinity Sample Problems Practice Problems

Limits at In…nity
Lecture Notes
page 1
Sample Problems
1. Compute each of the following limits.
a) lim 3x4
b)
2x5
c) lim
x!1
x!1
lim 3x4
d)
x! 1
x!1
2x5
lim
x! 1
2 6
x
3
e) lim
f)
g) lim 4x3
x!1
2 6
x
3
lim
x! 1
lim 4x3
h)
x! 1
2. Compute each of the following limits.
a) lim 2x
c) lim
x!1
b)
x!1
lim 2x
d)
x! 1
x
2
3
2
3
lim
x! 1
2x+3
x!1 3x+1
2x+3
f) lim x+1
x! 1 3
22x+1
x!1 3x 1
22x+1
h) lim
x! 1 3x 1
e) lim
x
g) lim
3. Compute each of the following limits.
a)
1
x!1 x
b)
lim
lim
d)
1
1x
x!
lim
x! 1
x!1
5
x!1 2x3
f)
7+
2x3 + 1
e) lim
c) lim
5
2x3
3x
lim
x! 1
8
x
g) lim
5 12
+
x x4
h) lim
5x3
x!1
5x3
i)
x
2x + 4
x3
x!1
2
2x + 4
x2
5x3
lim
x!1
2x + 4
x4
4. Compute each of the following limits.
a)
lim
x! 1
b) lim
x!1
2x5
2x5
8x4 + 7x3
8x4 + 7x3
10
c)
10
2x5 + 8x6
lim
x! 1
d) lim
x!1
2x5 + 8x6
5. Compute each of the following limits.
a)
x + x2 6
1 6x + 5x2 + 2x3
x2 + 9
x!1 5x + 2x2
3
lim
x!
b) lim
c)
x3
1 3x2
lim
x!
9x + 1
2x 15
Practice Problems
1. Compute each of the following limits.
a) lim
x!1
b)
lim
x! 1
3 15
x
8
3 15
x
8
1 8
x
x!1 3
c) lim
d)
lim
x! 1
c copyright Hidegkuti, Powell, 2009
1 8
x
3
e) lim 4x9
x!1
f)
lim 4x9
x! 1
g) lim
x!1
h)
lim
x! 1
7x10
7x10
Last revised: November 27, 2013
Limits at In…nity
Lecture Notes
page 2
2. Compute each of the following limits.
23x
x!1 5x
a) lim
b)
23x
1 5x
lim
x!
22x+3
1 5x 1
1
d)
1
lim
x!
f)
22x+3
1 4x 1
22x+3 3x
x!1
7x 2
lim
x!
1
g) lim
e) lim
c) lim
1
2x+3 3x
x!1
7x 2
22x+3
x!1 4x 1
22x+3
x!1 5x 1
1
1
h) lim
3. Compute each of the following limits.
3
x!1 x5
a) lim
b)
3
1 x5
x!
x!1
lim
x! 1
3+
5
x3
3+
x! 1
5x
j)
1
x!1
lim
x! 1
4x8 + x3
x4
x!1
3
p)
5
3x5 + 2x
x2
lim
o) lim
x
x! 1
l)
n)
3
5x
k) lim
7
6x
x!1
x
lim
3x5 + 2x
x2
m) lim
2
x+3
5x
x!1
7
6x
5
x3
lim
i) lim
2
5
+ 4
x 3x
1
x! 1
x!1
f)
2
5
+ 4
x 3x
1
lim
e) lim
h)
2
x+3
5x
x!1
lim
c) lim
d)
g) lim
lim
x! 1
5
x+7
4x8 + x3
x4
x+7
3x
2x
1
x! 1
3x
2x
4. Compute each of the following limits.
a)
lim
x! 1
b) lim
x!1
7x5 + x3
c)
7x5 + x3
lim
x! 1
d) lim
x!1
1 6
x
4
120x5
120x5
1 6
x
4
e)
f)
lim
x! 1
lim
x! 1
8x4
8x4
1
x+2
5
3x3
1
x+2
5
3x3
5. The graph of a polynomial function is shown on the picture below. What can we state about this polynomial
based on its end-behavior?
y
x
6. Compute each of the following limits.
a)
b)
1
1x
lim
x!
5
1 2x3
lim
x!
c)
lim
x! 1
d) lim
x!1
3
c copyright Hidegkuti, Powell, 2009
2
5
x3
2 11
+
x x4
2x2 + 3x + 1
x!1 3x2
5x + 2
e) lim
f)
lim
x! 1
3x3 + 2x + 1
5x 3
g)
h)
lim
x! 1
3x2
5x2
1
3x + 2
20x 2x2 42
1 5x3
20x2 105x
lim
x!
Last revised: November 27, 2013
Limits at In…nity
Lecture Notes
page 3
Sample Problems - Answers
1. a) 1
b) 1
2. a) 1
b) 0
c)
1
c) 0
3. a) 0
b) 0
4. a) 1
b)
5. a) 0
b)
d) 1
d) 1
c) 0
1
d)
c)
1
e) 0
7
c) 1
1
2
e)
f)
f) 1
e)
1
g) 1
g) 1
1
f) 3
h)
1
h) 0
g)
1
h)
5
i) 0
d) 1
1
Practice Problems - Answers
1. a)
1
2. a) 1
b) 0
3. a) 0
m)
b) 1
b) 0
1
b)
c) 1
1
lim f (x) =
x! 1
coe¢ cient.
6. a) 0
c) 0
n) 1
4. a) 1
5. Since
c) 1
b) 0
d) 1
d) 1
d) 1
o)
e) 32
e) 3
1
c)
e) 1
p)
1
f)
1
f) 32
f) 3
g) 1
1
e) 1
g)
g) 0
h)
1
h)
1
h) 1
1
i) 5
j) 5
k)
3
2
l)
3
2
1
d)
f) 1
1 and lim f (x) = 1, the polynomial is of odd degree and has a positive leading
x!1
c) 2
c copyright Hidegkuti, Powell, 2009
d) 3
e)
2
3
f)
1
g)
3
5
h) 0
Last revised: November 27, 2013
Limits at In…nity
Lecture Notes
page 4
Sample Problems - Solutions
1. Compute each of the following limits.
a) lim 3x4
x!1
Solution: Since the limit we are asked for is as x approaches in…nity, we should think of x as a very large
positive number. Then 3x4 is very large, and also positive because it is the product of …ve positive numbers.
3x4 =
3
x
positive
x
positive
positive
x
positive
x
positive
So the answer is 1. We state the answer: lim 3x4 = 1.
x!1
b)
lim 3x4
x! 1
Solution: Since the limit we are asked for is as x approaches negative in…nity, we should think of x as a very
large negative number. Then 3x4 is very large, and also positive because it is the product of one positive
and four negative numbers.
3x4 =
3
positive
x!1
x
negative
x
negative
x
negative
lim 3x4 = 1
So the answer is 1. We state the answer:
c) lim
x
negative
x! 1
2x5
Solution: Since the limit we are asked for is as x approaches in…nity, we should think of x as a very large
positive number. Then 2x5 is very large, and also negative because it is the product of one negative and
…ve positive numbers.
2x5 =
2
x
x
x
x
x
positive
negative
So the answer is
d)
lim
x! 1
positive
positive
2x5 =
1. We state the answer: lim
x!1
positive
positive
1
2x5
Solution: Since the limit we are asked for is as x approaches negative in…nity, we should think of x as a very
large negative number. Then 2x5 is very large, and also positive because it is the product of six negative
numbers.
2x5 =
2
x
x
x
x
x
negative
negative
So the answer is 1. We state the answer:
e) lim
x!1
negative
negative
negative
negative
2x5 = 1
lim
x! 1
2 6
x
3
Solution: Since the limit we are asked for is as x approaches in…nity, we should think of x as a very large
2 6
positive number. Then
x is very large, and also negative because it is the product of one negative and
3
six positive numbers.
2 6
x =
3
c copyright Hidegkuti, Powell, 2009
2
3
x
positive
x
positive
x
positive
x
positive
x
positive
x
positive
negative
Last revised: November 27, 2013
Limits at In…nity
Lecture Notes
So the answer is
1. We state the answer:
2 6
x
3
lim
x!1
=
page 5
1
2 6
x
x! 1
3
Solution: Since the limit we are asked for is as x approaches negative in…nity, we should think of x as a
2 6
x is very large, and also negative because it is the product of seven
very large negative number. Then
3
negative numbers.
2 6
2
x =
x
x
x
x
x
x
3
3 negative negative negative negative negative negative
f)
lim
negative
So the answer is
1. We state the answer:
2 6
x =
3
lim
x! 1
1
g) lim 4x3
x!1
Solution: Since the limit we are asked for is as x approaches in…nity, we should think of x as a very large
positive number. Then 4x3 is very large, and also positive because it is the product of four positive numbers.
4x3 =
4
positive
x
x
positive
x
positive
positive
So the answer is 1. We state the answer: lim 4x3 = 1
x!1
h)
lim 4x3
x! 1
Solution: Since the limit we are asked for is as x approaches negative in…nity, we should think of x as a very
large negative number. Then 4x3 is very large, and also negative because it is the product of one positive
and three negative numbers.
4x3 = 4
x
x
x
positive
So the answer is
1. We state the answer:
negative
lim
x! 1
4x3
negative
=
negative
1
2. Compute each of the following limits.
Let a > 0: Then the limit of the exponential function f (x) = ax is as follows.
lim ax = 1 and
Case 1.
If a > 1, then
Case 2.
If 0 < a < 1, then
lim ax = 0
x!1
x! 1
lim ax = 0 and
x!1
y
-5
-4
-3
-2
-1
0
1
lim ax = 1
x! 1
y
2
3
4
5
-5
-4
-3
-2
-1
0
1
x
a>1
c copyright Hidegkuti, Powell, 2009
2
3
4
5
x
0<a<1
Last revised: November 27, 2013
Limits at In…nity
Lecture Notes
a) lim 2x
and
x!1
page 6
lim 2x
b)
x! 1
Solution: Since 2 > 1; these limits are 1 and 0; i.e. lim 2x = 1 and lim 2x = 0.
x!1
c) lim
x!1
2
3
x
and
Solution: Since
d)
lim
x! 1
x! 1
x
2
3
2
< 1; these limits are 0 and 1, i.e. lim
x!1
3
2
3
x
= 0 and lim
x! 1
2x+3
e) lim x+1
x!1 3
Solution: We start by re-writing the exponential expressions.
there is only one exponential expression involving x:
2
3
x
f)
2x+3
1 3x+1
2x+3
2x+3
8
=
lim
= lim
1 3x+1
x! 1 3x+1
x! 1 3
2
3
2x+3
8
= lim
x+1
x!1 3
x!1 3
lim
8
lim
3 x!1
=
= 1.
2
3
x
x
2
3
=0
since
2
<1
3
lim
x!
Solution:
lim
x!
x
=
8
lim
3 x! 1
22x+1
g) lim x 1
x!1 3
Solution: We start by re-writing the exponential expressions.
there is only one exponential expression involving x:
2
3
x
=1
The goal is to bring it into a form where
x
22
2
22x+1
4x 6
22x 21
=
=
=
=6
x
1
3
3x 1
3x
x
3
3
31
Thus
h)
x
The goal is to bring it into a form where
2x 23
2x 8
8
2x+3
=
=
=
3x+1
3x 31
3x 3
3
Thus
2
3
22x+1
= lim 6
x!1 3x 1
x!1
lim
22x+1
1 3x 1
4
3
x
= 6 lim
x!1
4
3
4
3
x
x
=1
since
4
>1
3
lim
x!
Solution:
22x+1
= lim 6
1 3x 1
x! 1
lim
x!
4
3
x
= 6 lim
x! 1
4
3
x
=0
3. Compute each of the following limits.
1
a) lim
x!1 x
Solution: This is a very important limit. Since the limit we are asked for is as x approaches in…nity, we
should think of x as a very large positive number. The reciprocal of a very large positive number is a very
small positive number. This limit is 0.
1
x
Solution: Since the limit we are asked for is as x approaches negative in…nity, we should think of x as a very
large negative number. The reciprocal of a very large negative number is a very small negative number.
This limit is 0.
b)
lim
x! 1
c copyright Hidegkuti, Powell, 2009
Last revised: November 27, 2013
Limits at In…nity
Lecture Notes
page 7
5
2x3
Solution: Since the limit we are asked for is as x approaches in…nity, we should think of x as a very large
positive number. We divide 5 by a very large positive number. This limit is 0.
c) lim
x!1
d)
lim
x! 1
5
2x3
7+
8
x
Solution: This limit is 7 since the other two terms aproach zero as x approaches negative in…nity. Using
mathematical notation,
5
1 2x3
7+
lim
x!
x!1
7+0=
7
5 12
+
x x4
2x3 + 1
e) lim
8
5
8
= lim
+ lim
7 + lim
=0
3
x! 1
x! 1 x
x x! 1 2x
Solution: This limit is 1 since the …rst term approaches negative in…nity, the second term approaches 1
and the other two terms aproach zero as x approaches in…nity. Using mathematical notation,
lim
x!1
f)
lim
5 12
+
x x4
2x3 + 1
3x
5
x
2x3 + lim 1 + lim
= lim
x!1
x!1
x!1
12
x4
+ lim
x!1
=
1+1+0+0=
1
2
x
Solution: This problem is similar to the previous problems after a bit of algebra. We simply divide by x
and then the limit becomes familiar.
x! 1
lim
3x
g) lim
5x3
2
x
x! 1
= lim
x! 1
3x
x
2
x
= lim
x! 1
3
2
x
=3
2x + 4
x2
x!1
Solution:
5x3
lim
x2
x!1
h) lim
2x + 4
5x3
5x3
2x
4
2
4
+ 2 + 2 = lim
5x
+ 2
2
x!1
x!1
x
x
x
x x
2
4
= lim ( 5x) + lim
+ lim
= 1+0+0=
x!1
x!1
x!1
x
x2
=
lim
1
2x + 4
x3
x!1
Solution:
5x3
lim
x3
x!1
i)
lim
x!1
5x3
2x + 4
= lim
2x
4
5x3
+ 3 + 3
3
x
x
x
= lim
5
= lim
2x
4
5x3
+ 4 + 4
4
x
x
x
= lim
5
x
x!1
x!1
4
2
+ 3
2
x
x
=
5
2x + 4
x4
Solution:
lim
x!1
5x3
2x + 4
x4
c copyright Hidegkuti, Powell, 2009
x!1
x!1
2
4
+ 4
3
x
x
=0
Last revised: November 27, 2013
Limits at In…nity
Lecture Notes
page 8
4. Compute each of the following limits.
a)
lim
x! 1
2x5
8x4 + 7x3
10
Solution: The …rst term, 2x5 approaches in…nity and the secomd term, 8x4 approaches negative in…nity.
This does not give us enough information about the entire polynomial. A limit like this is called an
indeterminate. We will bring this expression to a form that is not an indeterminate. In this case,
factoring out the …rst term does the trick.
In case of a polynomial, the limits at in…nity and negative in…nity are
completely determined by its leading term. Recall that the leading term is the highest degree term.
2x5
lim
x! 1
8x4 + 7x3
2x5
10 = lim
x! 1
Here is the computation showing why this is true. We …rst factor out the entire leading term.
lim
x! 1
2x5
8x4 + 7x3
10
=
=
=
lim
2x5
lim
2x5
lim
2x5
x! 1
x! 1
x! 1
4
x
1+
lim
x! 1
7
5
+ 5
2
x
x
4
7
5
1+
+
x x2 x5
2x5
1 = lim
x! 1
We can now easily determine that this limit is 1. (See problem number 1.)
b) lim
x!1
2x5
8x4 + 7x3
10
Solution: In case of a polynomial, the limits at in…nity and negative in…nity are
completely determined by its leading term. Recall that the leading term is the highest degree term.
2x5
lim
x!1
8x4 + 7x3
2x5
10 = lim
x!1
Here is the computation showing why this is true. We …rst factor out the entire leading term.
lim
x!1
2x5
8x4 + 7x3
10
=
=
=
We can now easily determine that this limit is
c)
lim
x! 1
lim
2x5
lim
2x5
lim
2x5
x!1
x!1
x!1
4
x
1+
lim
x!1
7
5
+
2x2 x5
7
5
4
1+
+ 5
2
x 2x
x
2x5
1 = lim
x!1
1. (See problem number 1.)
2x5 + 8x6
Solution: In case of a polynomial, the limits at in…nity and negative in…nity are
completely determined by its leading term.
lim
2x5 + 8x6
=
lim
2x5 + 8x6
=
x! 1
x! 1
=
lim 8x6 = 1 because
x! 1
lim
8x6
lim
6
x! 1
x! 1
8x
2x5 = lim
x! 1
6
8x6
1
1
4x
= lim
x! 1
8x6
lim
x! 1
1
1
4x
1 = lim 8x
x! 1
We can now easily determine that this limit is 1. (See problem number 1.)
c copyright Hidegkuti, Powell, 2009
Last revised: November 27, 2013
Limits at In…nity
Lecture Notes
page 9
2x5 + 8x6
d) lim
x!1
Solution: In case of a polynomial, the limits at in…nity and negative in…nity are
completely determined by its leading term.
lim
2x5 + 8x6
=
lim
2x5 + 8x6
=
x!1
x!1
lim 8x6 = 1 because
x!1
lim 8x6
2x5 = lim 8x6
lim 8x6
1=1
x!1
=
x!1
1
4x
1
x!1
= lim 8x6
x!1
lim
x!1
1
1
4x
We can now easily determine that this limit is 1. (See problem number 1.)
5. Compute each of the following limits.
x + x2 6
=0
x! 1 6x + 5x2 + 2x3
Solution: The numerator approaches in…nity and the denominator approaches negative in…nity. This does
not give us enough information about the quotient. A limit like this is called an indeterminate. We
will bring this expression to a form that is not an indeterminate. Let us rearrange the polynomials in the
rational function given. Then we will factor out the leading term in the numerator and denominator.
a)
lim
x2 + x 6
lim
= lim
x! 1 2x3 + 5x2 + 6x
x! 1
1
6
x x2
5
3
1+
+ 2
2x x
x2 1 +
2x3
We now express the limit of the product as the product of two limits
1
6
x x2
5
3
1+
+ 2
2x x
x2 1 +
lim
x! 1
2x3
= lim
x! 1
1
6
x x2
5
3
1+
+ 2
2x x
1+
x2
2x3
lim
x! 1
The …rst expression can be simpli…ed and thus has a limit we can easily determine its limit. The second
expression, although looks unfriendly, is always going to approach 1.
x2
lim
x! 1 2x3
6
1
x x2
5
3
1+
+ 2
2x x
1+
lim
x! 1
1
1=0 1=0
1 2x
= lim
x!
The entire computation should look like this:
x2 + x 6
1 2x3 + 5x2 + 6x
lim
x!
=
=
c copyright Hidegkuti, Powell, 2009
1
6
x x2
5
3
1+
+ 2
2x x
x2 1 +
lim
x! 1
2x3
1
1=0 1=0
1 2x
x2
1 2x3
= lim
x!
1
6
1
1+
B
x x2 C
lim @
5
3 A
x! 1
1+
+ 2
2x x
0
lim
x!
Last revised: November 27, 2013
Limits at In…nity
Lecture Notes
page 10
x2 + 9
1
=
2
x!1 5x + 2x
3
2
Solution: Both numerator and denominator approach in…nity. This does not give us enough information
about the quotient. A limit like this is called an indeterminate. We will bring this expression to a form
that is not an indeterminate. Let us rearrange the polynomials in the rational function given. Then we will
factor out the leading term in the numerator and denominator.
b) lim
x2 1 +
x2 + 9
lim
x!1 2x2 + 5x
3
x3
x! 1 3x2
Solution:
c)
lim
lim
x! 1
9x + 1
=
2x 15
x3
3x2
9x + 1
2x 15
9
x2
1+
x2
= lim
x!1 2x2
lim
5
x!1
3
1+
2
2x
2x
9
1+ 2
1
1
1
x
= lim
lim
=
1=
5
3
x!1 2 x!1
2
2
1+
2x 2x2
=
lim
x!1
2x2
5
1+
2x
9
x2
3
2x2
1
x3 1
=
=
lim
x! 1
x3
1 3x2
lim
x!
3x2 1
9
+
x2
2
3x
1=
1
x3
5
x2
x3
= lim
x! 1 3x2
x
1 3
lim
x!
1=
1
lim
x! 1
1 1=
1
9
+
x2
2
3x
1
x3
5
x2
1
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c copyright Hidegkuti, Powell, 2009
Last revised: November 27, 2013