1. If f(2) = 5, f(3) = 2, f(4) = 5, g(2) = 6, g(3) = 2 and g(4) = 0,
find (f · g)(2) + f
g(3)
.
Solution. (f · g)(2) + f
g(3)
= f(2) · g(2) + f(2) = 5 · 6 + 5 = 35.
2. Evaluate the following limit
limx→0
2 −
√
4 − x
2
x
2
.
Solution.
limx→0
2 −
√
4 − x
2
x
2
= limx→0
2 −
√
4 − x
2
x
2
·
2 + √
4 − x
2
2 + √
4 − x
2
= limx→0
4 − (4 − x
2
)
x
2
(2 + √
4 − x
2
)
= limx→0
1
2 + √
4 − x
2
=
1
4
.
3. For which value of the constant c is the function f(x) continuous
on (−∞,∞)?
f(x) = (
c
2x − c x ≤ 1
cx − x x > 1.
Solution. The partial functions of f(x) are continuous for x < 1
and x > 1 because they are polynomials. To get f(x) continuous on
(−∞, ∞) we need
lim
x→1−
f(x) = lim
x→1+
f(x) = f(1).
This happens when c
2 − c = c − 1. Rearranging gives 0 = c
2 − 2c + 1 =
(c − 1)2
, and thus c = 1.
4. Compute
lim
x→π/2+
tan x.
Solution. From the graph of y = tan x, the limit is −∞. Or, since
tan x =
sin x
cos x
and sin(π/2) = 1 and cos(π/2) = 0, tan x has a vertical
asymptote at x = π/2. Thus the limit is either ∞ or −∞. For π/2 <
x < π, we have sin x > 0 and cos x < 0. Thus for x near π/2 but
greater than π/2, tan x < 0. Therefore the answer must be