Math 165 – Exam 1D - solutions
Problem 1 Evaluate the following limits.
2x + 4
12
a) lim
=
=4
x→4 x − 1
3
y+1
1
b) lim 2
=
y→−1 y + 5y + 4
3
cos2 (t) sin(4t) sin(2t)
8
=
2
t→0
9t
9
Solution In a), simply use substitution. In b), factor and cancel out y + 1
to get
1
y+1
=
2
y + 5y + 4
y+4
c) lim
the limit of which you can then compute again by substituting y = −1, since
the denominator is now nonzero.
In c), you need to split this up so you can use special trig limits, namely
8 cos2 (t) sin(4t) sin(2t)
cos2 (t) sin(4t) sin(2t)
=
·
·
9t2
9
4t
2t
Then the first factor has limit 8/9 and the second and third factors have
limit 1.
Problem 2 Evaluate the limits, if they exist.
4t2 − 5t + 8 cos t
4
=
2
t→∞
7t − 5t
7
3−z
= −∞
b) lim− 2
z→2 z − 4
Solution In a), we factor out the leading power t2 in numerator and denominator, getting
4 − 5t−1 + 8t−2 cos t
lim
t→∞
7 − 5t−1
and every term tends to zero except the 4 and the 7. In particular, we use
the Sandwich Theorem on
a) lim
−
8
8 cos t
8
≤
≤ 2
2
2
t
t
t
to see that the function in the middle has limit zero.
In b), the denominator has limit zero, but the numerator does not. Also, the
numerator is positive for z close to 2, and z 2 − 4 < 0 for z approaching 2
from the left. So the limit is negative infinity.
Problem 3 Evaluate the following limit.
√
√
x+4− 6
L = lim
x→2
x−2
Solution We cannot use substitution, since
and denomina√
√ both numerator
tor have limit 0. So we multiply both by x + 4 + 6, getting
x−2
√
√
x→1 (x − 2)( x + 4 +
6)
√
1
1
6
√ = √ =
.
= lim √
x→1
12
x+4+ 6
2 6
L = lim
Problem 4 Consider the function
f (x) = 2 + sin(5x).
a) Find the instantaneous rate of change of f (x) at x = 0. Use a limit
computation, no rules for derivatives that are not covered in Chapter 2.
b) Find an equation for the tangent line to the graph of f (x) at x = 0.
Solution
a) First, the instantaneous rate of change is
sin(5h)
1
= 5.
m = lim (2 + sin(5h) − (2 + 0)) = 5 lim
h→0
h→0 h
5h
b) The tangent line has equation y = 5x + 2.
Problem 5 Consider a function f (x) defined by


A
for x = 0,


B
for x = 3,
f (x) =
2
x
+
x
−
12



for all other values of x.
x2 − 3x
a) Can the constant A be chosen so that f (x) is continuous at x = 0, and if
so, what should it be?
b) Can the constant B be chosen so that f (x) is continuous at x = 3, and if
so, what should it be?
Solution a) No. The limit of f (x) at x = 0 does not exist, because
x2 + x − 12
x+4
=
2
x − 3x
x
and
lim+ f (x) = lim+
x→0
x→0
x+4
=∞
x
(you could also use that the left-handed limit is −∞).
b) Yes. We compute
x+4
7
= .
x→3
x
3
lim f (x) = lim
x→3
This is the value we need to choose for B to make f (x) continuous at 3.