MATH 147
Lab Key #3
2/9/2016
(1) Evaluate each limit:
2
(a) lim
−
x
−
4
x →4
9 − x2
(b) lim
x
x →3 3 −√
1 − 1 − x2
(c) lim
x →0
x2
Solution: (a)
lim
x →4−
(b)
2
2
1
= lim
= 2 · lim
= −∞
−
−
x−4
x
x →0
x →0 x
9 − x2
(3 + x ) (3 − x )
= lim
= lim (3 + x ) = 6.
3−x
x →3
x →3
x →3 3 − x
lim
(c)
lim
1−
x →0
√
1−
x2
= lim
x2
x →0
x2
1+
1+
x2
√
= lim
x →0
x →0
= lim
1−
1−
√
√
√
1 − x2 1 + 1 − x2
√
x2 1 + 1 − x2
√
1 − x2 − 1 − x2 − 1 − x2
√
x2 1 + 1 − x2
x2
= lim
x →0
1+
1
1
1
√ = .
√
=
2
1+ 1
1 − x2
1
(2) Consider the function f defined by
( 2
f (x) =
2x + x −6
x +2 ,
a,
x 6 = −2
x = 2.
What value should be assigned to a so that f ( x ) is continuous on R =
(−∞, ∞) .
Solution: Remember that a function f is continuous on R if and only if for each
b, the following equation is true:
lim f ( x ) = f (b) .
x →b
Therefore, we want
lim f ( x ) = f (−2) = a.
x →−2
Therefore, we need to find the above limit.
2x2 + x − 6
(2x − 3) ( x + 2)
= lim
= lim (2x − 3) = 2 (−2) − 3 = −7.
x+2
x+2
x →−2
x →−2
x →−2
lim
Hence, assigning the value a = −7, f is continuous on R.
(3) Suppose that the size of a population at time t ≥ 0 is given by
N (t) =
60
.
1 + 3e−t
(a) Determine the initial population size, N (0) .
(b) Determine the limiting population size, lim N (t) .
t→∞
Solution: (a) The initial population size is given by
N (0) =
60
60
60
=
=
= 15.
1+3
4
1 + 3e0
(b) The limiting population size is given by
60
60
=
= 60.
−
t
t→∞ 1 + 3e
1+0
lim N (t) = lim
t→∞
2
(4) Evaluate each limit:
x2 + 3
(a) lim
x →∞ 5x2 − 2x + 1
2x − 1
(b) lim
x →−∞ 3 − 4x
1 − 5x3
(c) lim
x →∞ 1 + 3x4
2x + x2
(d) lim
x →−∞ 3x + 1
Solution: (a)
1 + x32
x2 + 3
1+0
1
=
lim
=
= .
x →∞ 5x2 − 2x + 1
x →∞ 5 − 2 + 1
5
−
0
+
0
5
x
x2
lim
(b)
2− 1
2x − 1
2−0
1
= lim 3 x =
=− .
x →−∞ 3 − 4x
x →−∞
0
−
4
2
x −4
lim
(c)
1 − 5x3
= lim
x →∞
x →∞ 1 + 3x4
lim
1
x4
1
x4
−
5
x
+3
=
0−0
= 0.
0+3
(d) Using long division of polynomials, we can see that
x2 + 2x
1
5
5
= x+ −
.
3x + 1
3
9 9 (3x + 1)
∴ lim
x →−∞
x2 + 2x
5
1
5
5
= −∞ + − 0 = −∞.
= lim x + −
x →−∞ 3
3x + 1
9 9 (3x + 1)
9
3