Math 165 – Exam 1B - solutions
Problem 1 Evaluate the following limits.
5x + 2
22
a) lim
=
x→4 x − 1
3
cos2 (2y) sin(2y) sin(5y)
10
=
2
y→0
3y
3
b) lim
z−3
1
=
z→3 z 2 − 4z + 3
2
Solution In a), simply use substitution. In b), you need to split this up so
you can use special trig limits, namely
c) lim
cos2 (2y) sin(2y) sin(5y)
10 cos2 (2y) sin(2y) sin(5y)
=
·
·
3y 2
3
2y
5y
Then the first factor has limit 10/3 and the second and third factors have
limit 1.
In c), factor and cancel out z − 3 to get
z2
1
z−3
=
− 4z + 3
z−1
the limit of which you can then compute again by substituting z = 3, since
the denominator is now nonzero.
Problem 2 Evaluate the limits, if they exist.
5−s
1
a) lim− 2
=−
s→5 s − 25
10
2t2 + 5t − 40 sin t
2
=
2
t→∞
7t − 5t
7
Solution In a), the denominator and numerator has limit zero, so we can
cancel out s − 5 to get
−1
lim−
s→5 s + 5
and then use substitution s = 5. Approaching 5 from left or right does not
matter in this case.
b) lim
In b), we factor out the leading power t2 in numerator and denominator,
getting
2 + 5t−1 − 40t−2 sin t
lim
t→∞
7 − 5t−1
and every term tends to zero except the 2 and the 7. In particular, we use
the Sandwich Theorem on
−
40
40 sin t
40
≤
≤ 2
2
2
t
t
t
to see that the function in the middle has limit zero.
Problem 3 Evaluate the following limit.
√
√
x+2− 5
L = lim
x→3
x−3
Solution We cannot use substitution, since
and denomina√
√ both numerator
tor have limit 0. So we multiply both by x + 2 + 5, getting
x−3
√
√
x→3 (x − 3)( x + 2 +
5)
1
1
√ = √ .
= lim √
x→3
x+2+ 5
2 5
L = lim
Problem 4 Consider the function
f (x) = 4 + sin(3x).
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(3h)
m = lim (4 + sin(3h) − (4 + 0)) = 3 lim
= 3.
h→0 h
h→0
3h
b) The tangent line has equation y = 4x + 3.
Problem 5 Consider a function f (x)


A
for


B
for
f (x) =
2
x
+
4x


 2
for
x − 2x
defined by
x = 0,
x = 2,
all other values of x.
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 = 2, and if
so, what should it be?
Solution a) Yes. We compute
x2 + 4x
x+4
= lim
= −2.
2
x→0 x − 2x
x→0 x − 2
lim f (x) = lim
x→0
This is the value we need to choose for A to make f (x) continuous at 0.
b) No. The limit of f (x) at x = 2 does not exist, because
lim+ f (x) = lim+
x→2
x→2
x+4
= ∞.
x−2
(you could also use that the left-handed limit is −∞).