Calculus Volume 1

Chapter 4 | Applications of Derivatives

419

2 +2 x 4 x 3 −5 x +7

b. f ( x ) = 3 x

( Note: The degree of numerator is less than the degree of the denominator.)

2 +4 x

c. f ( x ) = 3 x

( Note: The degree of numerator is greater than the degree of the denominator.)

x +2

Solution a. The highest power of x in the denominator is x . Therefore, dividing the numerator and denominator by x and applying the algebraic limit laws, we see that lim x →±∞ = lim x →±∞ 3−1/ x 2+5/ x 3 x −1 2 x +5

lim x →±∞ (3−1/ lim x →±∞ (2+5/

x ) x )

=

lim x →±∞ 3− lim x →±∞ 1/ x lim x →±∞ 2+ lim x →±∞ 5/ x

=

3 2 .

= 3−0 2+0 =

f ( x ) = 3

Since lim

we know that y = 3 2

is a horizontal asymptote for this function as shown in

2 ,

x →±∞

the following graph.

Figure 4.53 The graph of this rational function approaches a horizontal asymptote as x →±∞. b. Since the largest power of x appearing in the denominator is x 3 , divide the numerator and denominator by x 3 . After doing so and applying algebraic limit laws, we obtain lim x →±∞ 3 x 2 +2 x 4 x 3 −5 x +7 = lim x →±∞ 3/ x +2/ x 2 4−5/ x 2 +7/ x 3 = 3 ( 0 ) +2 ( 0 ) 4−5 ( 0 ) +7 ( 0 ) =0. Therefore f has a horizontal asymptote of y =0 as shown in the following graph.

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