Math 113 Hw 5 Worksheet With Answers Page 2
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1
2
3
4
5which, using the Limit Law for quotients, is equal to
2
lim
3
3
=
.
5
4
lim
5 +
Therefore,
2
2
3y
3
lim
=
.
2
5y
+ 4y
5
4. Exercise 2.6.28. Find the limit
lim cos x.
Answer: Because cos(2nπ) = 1 for any n and because cos((2n + 1)π) =
1 for any n, this
limit cannot exist: no matter how far out we go on the x-axis, cos x is still oscillating between
1 and
1, so it never settles down to a limit.
5. Exercise 2.6.58.
(a) A tank contains 5000 L of pure water. Brine that contains 30 g of salt per liter of water
is pumped into the tank at a rate of 25 L/min. Show that the concentration of salt after
t minutes (in grams per liter) is
30t
C(t) =
.
200 + t
Answer: Since the tank starts with 5000 L of water and since 25 L/min of brine pours
in, the number of liters of water in the tank is given by
5000 + 25t.
On the other hand, the tank starts with no salt in it and, for each liter of brine that
pours in, 30 g of salt pours in. Hence, the number of grams of salt in the tank is given
by
25(30t).
Therefore, the concentration of salt is given by
25(30t)
25(30t)
30t
C(t) =
=
=
.
5000 + 25t
25(200 + t)
200 + t
(b) What happens to the concentration as t
?
Answer: As t
, the concentration of salt is given by
30t
lim C(t) = lim
.
200 + t
Dividing both numerator and denominator by t yields
1
30t
30
lim
= lim
= 30.
1
200
(200 + t)
+ 1
So eventually the concentration of salt in the tank approaches 30 g/L, which is the same
as the concentration of salt in the brine. In other words, the brine eventually overpowers
the pure water.
2
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