3 13 3 x f(x) = √₂² + 3x - 121 = 1² (√1 +²³2-¹) = (¹ + ¹ + 0 () − ¹) - 2 +0(1). |x| 1+ 1 x 1 2 x Therefore lim f(x) 448 3 x = lim x+∞ 2 lim f(x)= lim H118 312 418 2 x |x| X = lim x++∞ 2 312 lim 8118 312 = 312 (-1) = 32
3 13 3 x f(x) = √₂² + 3x - 121 = 1² (√1 +²³2-¹) = (¹ + ¹ + 0 () − ¹) - 2 +0(1). |x| 1+ 1 x 1 2 x Therefore lim f(x) 448 3 x = lim x+∞ 2 lim f(x)= lim H118 312 418 2 x |x| X = lim x++∞ 2 312 lim 8118 312 = 312 (-1) = 32
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.5: Rational Functions
Problem 54E
Related questions
Question
I don’t understand how the limits were found
![Exercise 1.
Consider the function f: dom f CR → R defined by f(x) = √x² + 3x − |x|.
(a) Determine the domain dom f, possible symmetry properties, the limits at the extreme points
of dom f and at too, and possible asymptotes.
The argument of the square root must be non negative, therefore the domain of the function
is
dom f = {x ER | x² + 3x = x(x + 3) ≥ 0} = (-∞, -3] U [0, ∞).
Since the domain of f isn't symmetric with respect to 0, the function is neither odd nor even.
In order to compute the required limits, observe that, as x→ ∞,
f(x)=√x² + 3x - |x| = |x|
Therefore
lim f(x) =
x4+∞
+
3
X
lim
x+∞
= x1 +
lim f(x) = lim
8118
3 x
2 x
3 |x|
xx 2 X
1 3
x
3
ت ات
lim
x+∞ 2
312
lim
xx 2
+o
=
312
(-1)
3
2
3 |x|
X
+0(1).
The function h(x)=√x² + 3x is the composition of the polynomial P(x) = x² + 3x (which is
continuous in R) with the function g(t) √t (continuous if t > 0), therefore it's continuous
on its domain. Since f is the sum of h with the continuous function -x], we have that fis
continuous on its domain. Therefore lim, f(x) = f(-3) = 0 and lim f(x) = f(0) =0. The
function f has no vertical asymptotes.
x-0
f has right horizontal asymptote y = 3/2 and left horizontal asymptote y = -3/2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2F130e915b-adfc-4c7c-8065-a5106e313766%2Fhlgwze8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Exercise 1.
Consider the function f: dom f CR → R defined by f(x) = √x² + 3x − |x|.
(a) Determine the domain dom f, possible symmetry properties, the limits at the extreme points
of dom f and at too, and possible asymptotes.
The argument of the square root must be non negative, therefore the domain of the function
is
dom f = {x ER | x² + 3x = x(x + 3) ≥ 0} = (-∞, -3] U [0, ∞).
Since the domain of f isn't symmetric with respect to 0, the function is neither odd nor even.
In order to compute the required limits, observe that, as x→ ∞,
f(x)=√x² + 3x - |x| = |x|
Therefore
lim f(x) =
x4+∞
+
3
X
lim
x+∞
= x1 +
lim f(x) = lim
8118
3 x
2 x
3 |x|
xx 2 X
1 3
x
3
ت ات
lim
x+∞ 2
312
lim
xx 2
+o
=
312
(-1)
3
2
3 |x|
X
+0(1).
The function h(x)=√x² + 3x is the composition of the polynomial P(x) = x² + 3x (which is
continuous in R) with the function g(t) √t (continuous if t > 0), therefore it's continuous
on its domain. Since f is the sum of h with the continuous function -x], we have that fis
continuous on its domain. Therefore lim, f(x) = f(-3) = 0 and lim f(x) = f(0) =0. The
function f has no vertical asymptotes.
x-0
f has right horizontal asymptote y = 3/2 and left horizontal asymptote y = -3/2.
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