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MAT 21A, Spring 2017
Solutions to homework Assignment 3
√
Section 2.5 38. (10 points) Find the limit limx→1 cos−1 (ln x).
√
√
√
Solution: As x approaches 1, x approaches 1 = 1, so ln( x) approaches ln(1) = 0,
and
√
π
lim cos−1 (ln x) = cos−1 (0) = .
x→1
2
√
Remark: Note that we used the fact that all three functions x, ln(x), cos−1 (x) are
continuous.
56. (10 points) Show that the function F (x) = (x − a)2 (x − b)2 + x takes on value
(a + b)/2 for some value of x.
Solution: Observe that F (a) = 0(a − b)2 + a = a, F (b) = b, and F (x) is continuous
on [a, b]. Therefore by the Intermediate Value Theorem it takes on all values between
F (a) = a and F (b) = b, in particular, (a + b)/2.
Section 2.6 31. (10 points) Compute the limit:
2x5/3 − x1/3 + 7
√ .
lim
x→∞ x8/5 + 3x +
x
5
1
Solution: We have 53 − 85 = 25
− 24
= 15
> 0, so x 3 is the highest power of x in this
15
15
fraction. Let us divide the numerator and the denominator by this power:
2x5/3 − x1/3 + 7
2 − x−4/3 + 7x−5/3
√
=
lim
.
x→∞ x8/5 + 3x +
x→∞ x−1/15 + 3x−2/3 + x1/2−5/3
x
lim
As x → ∞, the numerator approaches 2 − 0 + 0 = 2, and the denominator approaches
0 + 0 + 0 = 0. Therefore
2 − x−4/3 + 7x−5/3
= ∞.
x→∞ x−1/15 + 3x−2/3 + x1/2−5/3
2/3
x
110. (10 points) Consider the function y = 32 x−1
.
a) How does the graph behave as x → 0?
b) How does the graph behave as x → ±∞?
c) How does the graph behave near x = 1 and x = −1?
lim
x
Solution: The function is defined when x 6= 1 and x−1
≥ 0, so either x > 1 or x ≤ 0.
0
x
x
As x → 0, we have x−1 → −1 = 0, so f (x) → 0. As x → ±∞, we have x−1
→ 1, so
f (x) →
3 2/3 3
·1 = ,
2
2
x
so the graph has a horizontal asymptote y = 32 . At x → 1 we have x−1
→ ∞, so
f (x) → ∞, and f (x) has a vertical asymptote x = 1. Finally, f (x) is continuous ot
x = −1, and
2/3
3 1
3
lim f (x) = f (−1) =
= 5/3 .
x→−1
2 2
2
1
2
8
6
4
2
−4
−2
2
4
6
Remark: Some graphing calculators and programs may show that f (x) is defined for
0 < x < 1, since the cubic roots of negative numbers are well defined.