Math 165 - Practice Exam 1 - solutions
Problem 1 Evaluate the following limits.
a)
b)
c)
lim cos(π(x + 1)) = −1
x→0
tan(4y 2 )
4
=
2
y→0
3y
3
cos(2πt) − 1
lim
=0
t→1
t−1
lim
In c), you may use the addition formula of the cosine,
cos(a + b) = cos a cos b − sin a sin b
Solution In a), we only have to substitute x = 0.
In b), we use the intermediate variable u = 4y 2 , then the special trig limit
sin(u)/u, as in
sin u
sin(4y 2 )
= lim
= 1.
lim
2
u→0
y→0
4y
u
Then bring this into play via
(tan(4y 2 )
4 sin(4y 2 )
=
lim
y→0
y→0 3 · 4y 2 cos(4y 2 )
3y 2
4
4
= 1 · lim
= .
2
y→0 3 cos(4y )
3
lim
(the last limit can be done by substitution).
In c), we use the intermediate variable u = t − 1 and the addition formula of
the cosine, with a = 2πu and b = 2π, so
cos(2πt) = cos(2π(u + 1)) = cos(2πu)
and then the special trig limit for the cosine.
Problem 2 Evaluate the limits, if they exist.
a)
b)
c)
d)
e)
5
5t4 − 2t3 + 9
=
4
t→∞
2t + 100
2
3
3 − 2x + 4x
lim
=∞
x→∞
3x2 + 2x
5 − 2u
lim 2
does not exist
u→3 u − 9
5 − 2v
= −∞
lim+ 2
v→3 v − 9
2w4 − sin(w)
lim
= 0.
w→∞
w5 + 1
lim
Solution For a), we have a rational function with the same degree (four) in
numerator and denominator, so we can factor out t4 and then apply rules.
5 − 2t−1 + 9t−4
5
=
−4
t→∞
2 + 100t
2
lim
because all those powers of t with negative exponent have limit zero.
For b),
3 − 2x + 4x3
=∞
lim
x→∞
3x2 + 2x
because the numerator in this rational function has the higher degree (3
versus 2).
For c), the right-handed limit is −∞ and the left-handed limit is ∞ (similar
to d), see below). So the two-sided limit does not exist.
For d), the limit is −∞ because the limit of the numerator is −1 and the limit
of the denominator is 0, but the fraction is negative (numerator negative,
denominator positive) for v approaching 3 from the right.
For e), the limit is zero because this is a rational function with a numerator
of lower degree than the denominator (apart from the sin w term which is
dominated by any power of w).
Problem 3 Suppose the functions f (x) and g(x) are continuous everywhere and have these values: f (0) = 1, f (1) = 5, f (2) = 4, g(0) = 2,
g(2) = 1, g(4) = 3. Find limx→2 f (x) + g(x) and limx→2 f (g(x)).
Solution
lim g(x) + f (x) = 1 + 4 = 5
x→2
lim f (g(x)) = f (1) = 5
x→2
Problem 4 a) Find the instantaneous rate of change of
f (x) = 2x2 + 5x
at x = −1. 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). Make a sketch
of the graph and this tangent line, using the values
x −3 −2 −1 0 1
f (x)
3 −2 −3 0 7
Solution
a) First, the instantaneous rate of change is
1
m = lim (f (−1 + h) − f (−1))
h→0 h
1
= lim [2(1 − 2h + h2 ) − 5 + 5h − (−3)] = 1
h→0 h
b) Therefore the tangent line has equation y = m(x + 1) + f (−1) = x − 2.
Problem 5 Evaluate the following limit.
p
sin(4t2 )
lim+
t→0
t
Solution Take the square of that quantity! Then it is like something we
have done a number of times,
sin(4t2 )
=4
lim
t→0+
t2
Now the Main Limit theorem says we may take the square root of this again
which gives
p
sin(4t2 )
= 2.
lim+
t→0
t
Problem 6 Find the limit
√
√
x2 + x + 1 − x2 − 3x + 1
L = lim
.
x→0
x
Solution Substitution does not work, so we try algebra.
√
√
4x
√
x2 + x + 1 − x2 − 3x + 1 = √
2
x + x + 1 + x2 − 3x + 1
which we use for the limit in question,
L = lim √
x→0
4
√
= 2.
x2 + x + 1 + x2 − 3x + 1
The last limit can be done by substitution (Limit Law for quotients) since
the limit of the denominator is nonzero.
Problem 7 Consider the function f (x) defined below. What value can
be assigned to the constant a to make f (x) continuous everywhere?
2 + sinx x for x < 0,
f (x) =
x2 + a for x ≥ 0.
Solution a = 3 because the limit of f (x) for x approaching 0 from the left
is 3, from the right it is a, and the two need to agree.