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Determining Horizontal Asymptotes of Rational Functions, Exercises of Calculus

Iowa Wesleyan UniversityCalculus

How to find the horizontal asymptotes of a rational function by evaluating the limit of the function as x approaches infinity. It includes examples and general rules for polynomial functions.

Typology: Exercises

2021/2022

Uploaded on 09/12/2022

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.
HORIZONTAL ASYMPTOTES
In order to determine the horizontal asymptotes of a rational function R(x) = f(x)
g(x)we must evaluate
the following limit: lim
xโ†’โˆž
R(x).
Example: Evaluate lim
xโ†’โˆž
2x2+xโˆ’1
x2+ 1
Solution: Notice that 2x2+xโˆ’1โ†’ โˆž and x2+ 1 โ†’ โˆž as xโ†’ โˆž. For this function we can multiply
the numerator and denominator by 1
x2to determine the limit.
lim
xโ†’โˆž
2x2+xโˆ’1
x2+ 1 = lim
xโ†’โˆž
(2x2+xโˆ’1) (1
x2)
(x2+ 1) (1
x2)= lim
xโ†’โˆž
2 + 1
xโˆ’1
x2
1 + 1
x2
=
lim
xโ†’โˆž
2 + lim
xโ†’โˆž
1
xโˆ’lim
xโ†’โˆž
1
x2
lim
xโ†’โˆž
1 + lim
xโ†’โˆž
1
x2
=2+0โˆ’0
1+0 = 2
Here we are exploiting the fact that lim
xโ†’โˆž
1
xn= 0, where n > 0 is a rational number. Therefore
f(x) = 2x2+xโˆ’1
x2+ 1 has a horizontal asymptote along the line y= 2 as xapproaches โˆž.
Example: Find the horizontal asymptotes of g(x) = x+ 1
x2+ 3xโˆ’4
lim
xโ†’โˆž
x+ 1
x2+ 3xโˆ’4= lim
xโ†’โˆž
(x+ 1) (1
x2)
(x2+ 3x+ 4) (1
x2)= lim
xโ†’โˆž
1
x+1
x2
1 + 3
x2โˆ’4
x2
=
lim
xโ†’โˆž
1
x+ lim
xโ†’โˆž
1
x
lim
xโ†’โˆž
1 + lim
xโ†’โˆž
3
x2โˆ’lim
xโ†’โˆž
4
x2
=0+0
1 + 0 โˆ’0= 0
We conclude that g(x) has a horizontal asymptote along the line y= 0.
Note: In general for polynomial functions f(x) and g(x) we multiply the numerator and denominator
by 1
xmwhere mis the highest power of xthat appears in the denominator.
pf2

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HORIZONTAL ASYMPTOTES

In order to determine the horizontal asymptotes of a rational functionthe following limit: lim R(x) = f g^ ((xx)) we must evaluate xโ†’โˆž R(x). Example: Evaluate lim xโ†’โˆž^2 x^2 x^ + (^2) + 1^ x^ โˆ’^1 Solution: Notice that 2the numerator and denominator byx^2 + x โˆ’ 1 โ†’ โˆž 1 and x^2 + 1 โ†’ โˆž as x โ†’ โˆž. For this function we can multiply x^2 to determine the limit. xlimโ†’โˆž^2 x^2 x^ +^2 + 1^ x^ โˆ’ 1 = lim xโ†’โˆž^ (2x^2 +^ x^ โˆ’^ 1)

( (^) x 12 ) (x^2 + 1)^ (^ x^12 )^ = lim^ xโ†’โˆž

2 +^1 x โˆ’ (^) x^12 1 + x^12 =^ xlimโ†’โˆž^ 2 + lim x limโ†’โˆž^ x 1 + limโ†’โˆž x^1 xโ†’โˆž^ โˆ’^ x xlim 1 โ†’โˆž 2 x^12 = 2 + 01 + 0^ โˆ’ 0 = 2 Here we are exploiting the fact that (^) xlimโ†’โˆž x^1 n = 0, where n > 0 is a rational number. Therefore f (x) =^2 x^2 x^ + (^2) + 1^ x^ โˆ’ 1 has a horizontal asymptote along the line y = 2 as x approaches โˆž. Example: Find the horizontal asymptotes of g(x) = (^) x (^2) + 3x^ + 1x โˆ’ 4

xlimโ†’โˆž x^2 + 3^ x^ + 1x โˆ’ 4 = lim xโ†’โˆž^ (x^ + 1)

( (^) x 12 ) (x^2 + 3x + 4)^ (^ x^12 )^ = lim^ xโ†’โˆž

(^1) x + x (^12) 1 + x^32 โˆ’ (^) x^42 = x limโ†’โˆž 1 + lim^ xlimโ†’โˆž xโ†’โˆž^1 x^ + lim x 3 x 2 โ†’โˆž โˆ’ (^) x^1 xlimโ†’โˆž x 42 = (^) 1 + 00 + 0 โˆ’ 0 = 0 We conclude that g(x) has a horizontal asymptote along the line y = 0. Noteby 1 : In general for polynomial functions f (x) and g(x) we multiply the numerator and denominator xm^ where^ m^ is the highest power of^ x^ that appears in the denominator.

Given the rational function R(x) = f g^ ((xx) )= (^) bamnxxmn^ ++^ abnmโˆ’โˆ’^11 xxnmโˆ’โˆ’^1 1 + (^) +^ ยท ยท ยท ยท ยท ยท^ + +^ a b^11 xx^ + +^ a b^00 , the horizontal asymptotes for R(x) are determined by the following cases: (i) If deg(f (x)) > deg(g(x)), R(x) has no horizontal asymptote. (ii) If deg(f (x)) = deg(g(x)), R(x) has a horizontal asymptote at y = (^) bamn. (iii) If deg(f (x)) < deg(g(x)), R(x) has a horizontal asymptote at y = 0. Note: deg(f (x)) and deg(g(x)) represent the degree of f (x) and g(x), respectively. Examples (i) R(x) = x x^32 + 1โˆ’ 1 , deg(f (x)) > deg(g(x)). R(x) has no horizontal asymptote.

(ii) y = R 4 (x) =^4 x^22 โˆ’x (^2 3) + 7x^ + 1 , deg(f (x)) = deg(g(x)). R(x) has an horizontal asymptote along the line 2 = 2. (iii) R(x) = (^) xx (^2) + 5+ 1x, deg(f (x)) < deg(g(x)). R(x) has an horizontal asymptote along the line y = 0.