Limits Involving Radical
Functions
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Printed: December 16, 2014
AUTHOR
CK-12
www.ck12.org
C HAPTER
Chapter 1. Limits Involving Radical Functions
1
Limits Involving Radical
Functions
Objectives
In this concept, you will learn various ways to evaluate limits involving radical functions, such as direct substitution
and transformations to indeterminate or undefined forms.
Concept
There are many problems that will involve taking the nth root of a variable expression, so it is natural that there may
sometimes be a need to find the limit of a function involving radical expressions, using square or cube roots, or other
roots. Do you think that finding the limit of a function involving radicals would be any different than finding the
limit of polynomial or rational functions? Can you think of any ways that radicals might present different problems
than polynomials?
Watch This
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/79184
http://www.youtube.com/watch?v=5meojrrJM0U - James Sousa: Ex: Limits at Infinity of a Function Involving a
Square Root
Guidance
Whenever possible, use direct substitution to see if a limit can be evaluated. If not, other methods to evaluate the
limit need to be explored.
Example A
Find the following limits for the function f (x) =
√
a. limx→9 x − 3
√
b. limx→∞ x − 3
√
x − 3:
Solution:
a. The following steps
1
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√
√
x − 3 = lim x − lim 3
x→9
x→9
x→9
√
√
lim x − 3 = lim x − lim 3
x→9
x→9
x→9
√
= 9−3
lim
=0
√
Therefore, limx→9 x − 3 = 0, which could have been determined by directly evaluating f (x) at x = 9, i.e., by using
direct substitution.
b. Evaluating
f (x) at ever increasing positive values of x shows that f (x) increase without bound. Therefore,
√
limx→∞ x − 3 = ∞.
In both of the above cases, direct substitution could be used to evaluate the limits.
Example B
√
x2 + 3 , find the following:
Given g(x) = 7x+5
√
x2 + 3
a. limx→∞ 7x+5
√
x2 + 3
b. limx→−∞ 7x+5
Solution:
a. First we notice that we should exclude x = − 75 in any evaluation. Using direct substitution to find the limit results
in the indeterminate form ∞
∞ . To transform the radical expression to a better form, use the fact that the value of x is
going to larger and larger positive values. This allows the following:
√
x2 + 3
= lim
lim
x→∞
x→∞ 7x + 5
q
x2 1 + x32
x 7 + 5x
q
√
x2 1 + x32
2
x +3
lim
= lim
x→∞ 7x + 5
x→∞
x 7 + 5x
q
|x|
1 + x32
= lim
x→∞
x 7 + 5x
q
lim
1 + x32
= x→∞
lim 7 + 5x
x→∞
1
=
7
Therefore, limx→∞
√
x2 + 3 = 1 .
7x+5
7
b. The solution to evaluating the limit at negative infinity is similar to the above approach except that x is always
negative.
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Chapter 1. Limits Involving Radical Functions
√
x2 + 3
lim
= lim
x→−∞ 7x + 5
x→−∞
q
x2 1 + x32
x 7 + 5x
q
|x|
1 + x32
= lim
x→−∞
x 7 + 5x
q
1 + x32
lim
x→−∞
. . . Note the denominator has a - because x < 0
=
− lim 7 + 5x
x→−∞
1
=−
7
Therefore, limx→−∞
√
x2 + 3 = − 1 .
7x+5
7
Again, a similar procedure was used to evaluate some of the rational functions in the previous concept.
Example C
Given the function h(x) =
√
x−3
x−9
, find
a. limx→9 h(x)
b. limx→∞ h(x)
c. limx→−∞ h(x)
Solution:
a. Using direct substitution to find the limit results in the indeterminate form 00 . In order to evaluate the limit, we
need to transform the expression to remove the indeterminate form. This is accomplished by using the relationship
for the difference of squares of real numbers: x2 − y2 = (x + y)(x − y).
We then rewrite and simplify the original function as follows:
lim
x→9
√
√
x−3
x−3
√
= lim √
. . . Use the difference of squares factoring to remove the 0 in the denominator.
x→9 ( x − 3) ( x + 3)
x−9
1
= lim √
x→9 ( x + 3)
1
√
=
limx→9
9+3
1
=
6
Hence limx→9
√
x−3
x−9
= 16 .
b. As x increases to large positive values, the function takes on the indeterminate form ∞
∞ . The transformation above
can also be used to evaluate the limit (Approach 1), as well as the technique used in evaluating rational functions
(Approach 2).
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Approach 1
√
√
x−3
x−3
√
lim
= lim √
x→∞ x − 9
x→∞ ( x − 3)( x + 3)
1
= lim √
x→∞ ( x + 3)
1
√
=
lim ( x + 3)
Approach 2
√ x 1 − √3x
= lim
x→∞
x 1 − 9x
3
√
1
−
1
x
= lim √ · lim
9
x→∞
x x→∞ 1 − x
= 0·1
x→∞
=0
Hence limx→∞
√
x−3
x−9
=0
= 0.
c. The solution to this problem is that limx→−∞ h(x) does not exist because the domain of h(x) does not include
x < 0.
Concept Question Wrap-up
You will recall the question at the beginning of the lesson:
“Do you think that finding the limit of a function involving radicals would be any different than finding the
limit of polynomial or rational functions?
Can you think of any ways that radicals might present different problems than polynomials?”
The examples in this section show that some of the methods for evaluating limits involving polynomials and rational
functions can be used to find the limits of radical functions. The use of direct substitution is a common method.
Transforming indeterminate or undefined forms by finding and canceling common factors in the numerator and
denominator, or factoring and simplifying the highest degree powers of variables represent common approaches.
One of the noteworthy differences between polynomial and radical functions is that the domain
√ of polynomials can
include all real values of the independent variable, but the domain of radical functions, e.g., x, is restricted.
Vocabulary
Radical functions are functions which contain nth roots of variable expressions.
Guided Practice
Find limx→0
√
x + 4−2 .
x
Solution:
Using direct substitution to find the limit of the function results in the indeterminate form 00 . To transform the radical
expression to a better form, do the following:
4
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Chapter 1. Limits Involving Radical Functions
√
√
√
x+4−2
x+4−2
x+4+2
= lim
· √
lim
x→0
x→0
x
x
x+4+2
x+4−4
√
= lim
x→0 x ·
x+4+2
x
√
= lim
x→0 x ·
x+4+2
1
= lim √
x→0
x+4+2
1
=
4
Therefore, limx→0
. . . Rationalize the numerator: multiply by
the conjugate of the numerator.
√
x + 4−2 = 1 .
x
4
Practice
Find each of the following limits if they exist.
√
1. limx→3 √x
2. limx→8 √x − 7
x−2
3. limx→4 x−4
√
+ 3−2
4. limx→1 xx−1
√
x
5. limx→0+ p
√
1
+
x−1 p
6. limx→∞
x2 − 5x − x
√
6 + 3x2 + 1
7. limx→∞ x 4x
3 +3
√
√
x
+
5
− 5
8. limx→0 p x
x2 + 4x
9. limx→3
3
10. limx→−1√
(x2 + 2x + 10) 2
+ 5−3
11. limx→4 xx−4
p
12. limx→1
2x3 + 3x2 + 7
p
3
13. limx→3
2x2 − 10
14. limx→7 √ 5x
x√+ 2
7− x2 + 49
15. limx→0
x
5