MATH 241
HOMEWORK 4 SOLUTIONS
H4P1.
√
3
x−1
(i) lim √
x→1
x−1
√
√
√
√
√
3
3
3
x−1
x−1
x+1
x2 + 3 x + 1
lim √
= lim √
·√
· √
√
x→1
x − 1 x→1 x − 1
x + 1 3 x2 + 3 x + 1
√
( x + 1)(x − 1)
√
= lim
√
x→1 (x − 1)( 3 x2 + 3 x + 1)
√
x+1
= lim √
√
x→1 3 x2 + 3 x + 1
2
= .
3
√
√
√
Another solution would be to let t = 6 x. Then t2 = 3 x, t3 = x,
and t → 1 as x → 1. We have
√
3
t2 − 1
x−1
= lim 3
lim √
x→1
x − 1 t→1 t − 1
(t + 1)(t − 1)
= lim
t→1 (t − 1)(t2 + t + 1)
t+1
= lim 2
t→1 t + t + 1
2
= .
3
xn − an
x→a x − a
n
n
(x − a) xn−1 + xn−2 a + xn−3 a2 + · · · + x2 an−3 + xan−2 + an−1
x −a
lim
= lim
x→a x − a
x→a
x−a
= lim xn−1 + xn−2 a + xn−3 a2 + · · · + x2 an−3 + xan−2 + an−1
(ii) lim
x→a
n−1
=a
+ an−2 a + an−3 a2 + · · · + a2 an−3 + aan−2 + an−1
= an−1 + an−1 + an−1 + · · · + an−1 + an−1 + an−1
= nan−1 .
1
2
MATH 241
HOMEWORK 4 SOLUTIONS
H4P2. (a) Let P0 be a point on the equator, and Pθ be a point obtained from P0
by rotating it around the circle by angle θ counterclockwise. Let T (θ) be
the temperature at Pθ . We assume that T (θ) is continuous on [0, 2π] and
T (0) = T (2π). Consider f (θ) = T (θ) − T (θ + π) for θ in [0, π], so that f
is the difference in temperature for two points opposite each other on the
circle.
If f (0) = 0, then T (0) = T (π) and we have found two diametrically
opposite points with the same temperature.
If f (0) 6= 0, then notice that
f (π) = T (π) − T (2π)
= T (π) − T (0)
= −(T (0) − T (π))
= −f (0).
So if f (0) is positive, then f (π) is negative (and vice versa). Since f (θ)
changes sign between θ = 0 and θ = π, the Intermediate Value Theorem
tells us that there is a point c, 0 < c < π, where f (c) = 0. This means that
T (c) = T (c + π) and so c and c + π are diametrically opposite points with
the same temperature.
(b) We can apply the method above to any circle on the surface of the earth.
The solution to part (a) is not dependent on the fact that we were considering the equator, just that we were considering a circle.
H4P3.
(i)
x2 − 25
>0
(x − 2)(x2 − x + 6)
x2 − 25
Let f (x) =
. Setting the denominator equal to
(x − 2)(x2 − x + 6)
zero, we have that x − 2 = 0 or x2 − x + 6 = 0. Since x2 − x + 6 = 0 has
no real solutions, we conclude f is not continuous for x = 2. Setting
the numerator equal to zero, we have that x2 − 25 = (x+ 5)(x − 5) = 0,
so x = ±5. Thus we have the intervals
(−∞, −5), (−5, 2), (2, 5), and (5, ∞).
For x ∈ (−∞, −5), f (x) < 0. For x ∈ (−5, 2), f (x) > 0. For x ∈ (2, 5),
f (x) < 0. Finally, for x ∈ (5, ∞), f (x) > 0. Consequently, the solution
set is
(−5, 2) ∪ (5, ∞).
(ii)
x−4
≤0
(x − 2)(x2 − 1)
x−4
Let f (x) =
. Setting the denominator equal to zero,
(x − 2)(x2 − 1)
we have that x − 2 = 0 or x2 − 1 = (x + 1)(x − 1) = 0. We conclude f
is not continuous only for x = 2 and x = ±1. Setting the numerator
MATH 241
HOMEWORK 4 SOLUTIONS
3
equal to zero, we have that x − 4 = 0, so x = 4. Thus we have the
intervals
(−∞, −1), (−1, 1), (1, 2), (2, 4), and (4, ∞).
For x ∈ (−∞, −1), f (x) > 0. For x ∈ (−1, 1), f (x) < 0. For x ∈ (1, 2),
f (x) > 0. For x ∈ (2, 4), f (x) < 0. Finally, for x ∈ (4, ∞), f (x) > 0.
Consequently, the solution set is
(−1, 1) ∪ (2, 4].
(iii) |x2 − 4x| ≥ 6 Let f (x) = |x2 − 4x| − 6. Then f (x) = 0 if and only
2
2
if |x2 − 4x| = 6. This is equivalent
√ to x − 4x = 6 or x − 4x = −6.
The first equation gives x = 2 ± 10, and the second equation has no
real roots. Note that f is continuous for all x, as a composition of a
polynomial function and absolute value function. Therefore we obtain
three intervals:
√
√
√
√
(−∞, 2 − 10), (2 − 10, 2 + 10), and (2 + 10, ∞).
As f (−10) = 134 > 0, f (0) = −6 < 0, and f (10) = 54 > 0, f (x)
is positive on the first and the third intervals, and is negative on the
second interval. Consequently, our solution set is
√
√
(−∞, 2 − 10] ∪ [2 + 10, ∞).
Another solution is as follows. The inequality is equivalent to the
following statement:
x2 − 4x ≥ 6
or x2 − 4x ≤ −6.
Consider the first case: x2 − 4x ≥ 6. Let f (x) = x2 − 4x − 6. Setting
f (x) = 0 and solving for x yields
√
√
√
4 ± 16 + 24
4 ± 40
x=
=
= 2 ± 10.
2
2
2
Now consider the second case: x − 4x ≤ −6. Let g(x) = x2 − 4x + 6.
Setting g(x) = 0 and solving for x yields
√
4 ± 16 − 24
x=
,
2
which does not yield real values of x. Our intervals include
√
√
√
√
(−∞, 2 − 10), (2 − 10, 2 + 10), and (2 + 10, ∞).
√
√
√
For x ∈ (−∞, 2− 10),
√f (x) > 0. For x ∈ (2− 10, 2+ 10), f (x) < 0.
Finally, for x ∈ (2 + 10, ∞), f (x) > 0. Consequently, our solution
set is
√
√
(−∞, 2 − 10] ∪ [2 + 10, ∞).
4
MATH 241
(iv)
HOMEWORK 4 SOLUTIONS
x−1
x+1
>
x+2
x+5
x−1 x+1
x−7
Let f (x) =
−
=
. Setting the denominator
x+2 x+5
(x + 5)(x + 2)
equal to zero, we conclude that f is not continuous for x = −5 and
x = −2. Setting the numerator equal to zero, we have that x − 7 = 0
or x = 7. Thus our intervals are
(−∞, −5), (−5, −2), (−2, 7), and (7, ∞).
For x ∈ (−∞, −5), f (x) < 0. For x ∈ (−5, −2), f (x) > 0. For
x ∈ (−2, 7), f (x) < 0. Finally, for x ∈ (7, ∞), f (x) > 0. Consequently,
our solution set is
(−5, −2) ∪ (7, ∞).