410
Chapter 4 | Applications of Derivatives
Solution
a. Using
the
algebraic
limit
laws,
we
have
⎛ ⎝ 5− 2 x 2
⎞ ⎠ = lim x →∞ 5−2 ⎛
⎞ ⎠ .
⎛ ⎝ lim x →∞
⎞ ⎠ =5−2·0=5.
1 x
1 x
lim x →∞
⎝ lim x →∞
f ( x ) =5. Therefore, f ( x ) =5− 2 x 2
lim x
has a horizontal asymptote of y =5 and f
Similarly,
→ -∞
approaches this horizontal asymptote as x →±∞ as shown in the following graph.
Figure 4.43 This function approaches a horizontal asymptote as x →±∞.
b. Since −1≤ sin x ≤1 for all x , we have
x x ≤ 1 x
−1
x ≤ sin
for all x ≠0. Also, since
−1 x =0= lim x →∞ 1 x ,
lim x →∞
we can apply the squeeze theorem to conclude that lim x →∞ sin x x =0.
Similarly,
sin x
lim x →−∞
x =0.
Thus, f ( x ) = sin x
x has a horizontal asymptote of y =0 and f ( x ) approaches this horizontal asymptote
as x →±∞ as shown in the following graph.
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