Math 160 - Assignment 2 Solutions - Summer 2012 - BSU - Jaimos F Skriletz
1
1. Directional Limits

4 − x2



x
Consider the piecewise function f (x) =
 1


6−x
x≤1
1<x<3
x=3
x>3
(a)
lim f (x) = 4 − (1)2 = 3
x→1−
lim f (x) = (1) = 1
x→1+
lim f (x) = D.N.E because lim− f (x) 6= lim+ f (x)
x→1
x→1
x→1
2
f (1) = 4 − (1) = 3
(b)
lim f (x) = (3) = 3
x→3−
lim f (x) = 6 − (3) = 3
x→1+
lim f (x) = 3 because lim− f (x) = lim+ f (x) = 3
x→1
x→1
x→1
f (1) = 1
(c) The graph of f (x) is
2. Limits
Calculate out the following limits:
3x2 + 2x − 1
3(1)2 + 2(1) − 1
4
2
=
= =
2
x→1 x + 3x + 2
(1)2 + 3(1) + 2
6
3
(a) lim
3x2 + 2x − 1
(3x − 1)(x + 1)
3x − 1
3(−1) − 1
−4
= lim
= lim
=
=
= −4
x→−1 x2 + 3x + 2
x→−1 (x + 2)(x + 1)
x→−1 x + 2
(−1) + 2
1
(b) lim
[2(x + h)2 − 3(x + h)] − [2x2 − 3x]
2(x2 + 2xh + h2 ) − 3x − 3h − 2x2 + 3x
4xh + 2h2 − 3h
= lim
= lim
=
h→0
h→0
h→0
h
h
h
h(4x + 2h − 3)
lim
= lim (4x + 2h − 3) = 4x + 2(0) − 3 = 4x − 3
h→0
h→0
h
(c) lim
Math 160 - Assignment 2 Solutions - Summer 2012 - BSU - Jaimos F Skriletz
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3. Infinite Limits/Limits at Infinity
Calculate the following limits that include Infinity:
2x2 − 3x − 2
2(−2)2 − 3(−2) − 2
12
=
=
x→−2 x2 − x − 6
(−2)2 − (−2) − 6
0
This is a infinite limit form, so this function has a vertical asymptote at x = −2. So we can say that the limit
does not exist or is infinite. Further more we can look at the sign of the expression on each side of x = −2 to say
the following with directional limits
(a) lim
lim −
x→−2
2x2 − 3x − 2
= +∞
x2 − x − 6
lim +
x→−2
2x2 − 3x − 2
= −∞
x2 − x − 6
2x2 − 3x − 2
2(3)2 − 3(3) − 2
7
=
=
2
x→3 x − x − 6
32 − 3 − 6
0
This too is an infinite limit form, so this function also has a vertical asymptote at x = 3. Looking at the sign at
each side of x = 3 we can then say that
(b) lim
lim−
x→3
(c)
2x2 − 3x − 2
= −∞
x2 − x − 6
lim+
x→3
2x2 − 3x − 2
= +∞
x2 − x − 6
2 − x3 − x22
(2x2 − 3x − 2) x12
2−0−0
2x2 − 3x − 2
=
lim
=
=
lim
=2
x→±∞ 1 − 1 − 62
x→±∞ x2 − x − 6
x→±∞ (x2 − x − 6) 12
1−0−0
x
x
x
Thus this rational function has a horizontal asymptote at the line y = 2
lim
(d) Optional : The above three limits tell us that the rational function
R(x) =
2x2 − 3x − 2
x2 − x − 6
has vertical asymptote at x = −2 and x = 3 along with a horizontal asymptote at y = 3. You could have used your
graphing calculator to graph this function to understand the limits better. Thus you should look at the following
graph of R(x) and try to better understand the behavior the limits are expressing.
Math 160 - Assignment 2 Solutions - Summer 2012 - BSU - Jaimos F Skriletz
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4. Revenue, Costs and Profit
The owner of a 100 unit apartment building knows that he can fill all 100 units if he charges $460 per month. Research
shows that for each $10 per month increase in the rent, that one less apartment unit is filled. Further the average cost
of upkeep for an apartment is $75 per month for an occupied unit and $15 per month for an empty unit.
Let x be the number apartments rented at p dollars per month.
(a) If the demand is linear then it passes though the point
(x1 , p1 ) = (100, 460)
and has slope
m=
∆p
+10
=
= −10
∆x
−1
Thus the equation of the line is
p − p1 = m(x − x1 )
p − 460 = −10(x − 100)
p = −10x + 1460
(b) The total Revenue for renting x units at p per unit is
R(x) = xp = x(−10x + 1460) = −10x2 + 1460x
(c) The Cost to upkeep the apartments is
C(x) = 75x + 15(100 − x) = 75x + 1500 − 15x = 60x + 1500
Thus the Profit for renting these apartments is
P (x) = R(x) − C(x) = [−10x2 + 1460x] − [60x + 1500] = −10x2 + 1400x − 1500
(d) The break even points occur when
R(x) = C(x)
2
−10x + 1460x = 60x + 1500
2
−10x + 1400x − 1500 = 0
x=
−b ±
√
√
−1400 ± 190000
b2 − 4ac
=
≈ 1 or 139 units
2a
−20
(Note I should have included some large fixed cost for the mortgage/tax on the apartments to make more realistic
break even points.)