Math 165 – Exam 1A – solutions
Problem 1 Evaluate the following limits.
4x + 2
22
11
a) lim
=
=
x→5 x + 1
6
3
y−2
1
b) lim 2
=
y→2 y + 3y − 10
7
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 − 2
to get
y−2
1
=
2
y + 3y − 10
y+5
c) lim
the limit of which you can then compute again by substituting y = 2, 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.
5
5t2 + 2t + 8 sin t
=
2
t→∞
9t − 5t
9
3−z
b) lim+ 2
=∞
z→2 z − 4
Solution In a), we factor out the leading power t2 in numerator and denominator, getting
5 + 2t−1 + 8t−2 sin t
lim
t→∞
9 − 5t−1
and every term tends to zero except the 5 and the 9. In particular, we use
the Sandwich Theorem on
a) lim
−
8
8 sin 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 > 2. So the limit
is positive infinity.
Problem 3 Evaluate the following limit.
√
√
x+2− 3
L = lim
x→1
x−1
Solution We cannot use substitution, since
and denomina√
√ both numerator
tor have limit 0. So we multiply both by x + 2 + 3, getting
x−1
√
√
x→1 (x − 1)( x + 2 +
3)
√
1
1
3
√ = √ =
.
= lim √
x→1
6
x+2+ 3
2 3
L = lim
Problem 4 Consider the function
f (x) = 3 + sin(4x).
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
1
sin(4h)
m = lim (3 + sin(4h) − (3 + 0)) = 4 lim
= 4.
h→0 h
h→0
4h
b) The tangent line has equation y = 4x + 3.
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.