Math 165 Section A
Professor Lieberman
September 3, 2004
SOLUTIONS TO FIRST IN-CLASS EXAM
1. The function
2u + 7
√
(u − 5) u2 + 1
is a combination of rational functions and a square root so it’s continuous everywhere
that it’s defined, and it’s undefined only when the denominator is zero, which is
u=5 .
f (u) =
2. The fraction
2x + 4
√
x2 + 5
is defined everywhere, so there are no vertical asymptotes.
To find the horizontal asymptotes, we must compute two limits:
2+ 4
2x + 4
lim √
= lim q x = 2
x→∞
x2 + 5 x→∞ 1 + 52
since
√
x
x2 = x for x > 0 and
4
−2 + −x
2x + 4
lim √
= lim q
= −2
x→−∞
x2 + 5 x→∞ 1 + 52
because
√
x
x2 = −x if x < 0. Therefore the horizontal asymptotes are y = 2, y = −2 .
3. (a) This has the form 00 , so
1
sin(3t) + 4t
lim
lim
t→0 sec t
t→0
t
sin(3t) 4t
= lim cos t lim
+
t→0
t→0
t
t
sin(3t) + 4t
lim
=
t→0
t sec t
= (1)(3 + 4) = 7 .
(b) Again, this has the form 00 , so
2x2 − 6xπ + 4π 2
(2x − 4π)(x − π)
= lim
2
2
x→π
x→π (x + π)(x − π)
x −π
2x − 4π
−2π
= lim
=
= -1 .
x→π x + π
2π
lim
(c) Now we note that x/|x| = −1 for x < 0, so
x
= -1 .
lim−
x→0 |x|
4. For the limit to exist, we need the one-sided limits to agree, so
c(3)2 + 1 = c(3) − 2
or 9c + 1 = 3c − 2. This algebraic equation has the solution c = −1/2 .