Answer Key for Exam 1
1(a) lim
x→3
x2−2x−3
x2−5x+ 6 =32−2·3−3
32−5·3+6 =0
0, so we have to do something. In particular, there must be factors
of x−3 on top and bottom that are causing these zeros. So we write
lim
x→3
x2−2x−3
x2−5x+ 6 = lim
x→3
(x+ 1)(x−3)
(x−2)(x−3) = lim
x→3
x+ 1
x−2=3 + 1
3−2= 4.
1(b) lim
h→0
√4+9h−√4−3h
his also 0
0, so we have to do something, namely multiply by the conjugate:
lim
h→0
√4+9h−√4−3h
h= lim
h→0
√4+9h−√4−3h
h
√4 + 9h+√4−3h
√4 + 9h+√4−3h
= lim
h→0
4+9h−(4 −3h)
h
1
√4+9h+√4−3h
= lim
h→0
9h+ 3h
h
1
√4+9h+√4−3h
= lim
h→0
12
√4+9h+√4−3h
=12
√4 + √4=12
2+2 = 3.
2. If f(x) =
x2+ 3x−4 if x≤0
x2−4 if 0 < x ≤2
4x−8 if x≥2.
then
(a) lim
x→0−f(x) = lim
x→0−¡x2+ 3x−4¢= 02+ 3 ·0−4 = −4.
(b) lim
x→0+ f(x) = lim
x→0+ ¡x2−4¢= 02−4 = −4. Since we got the same answer in (a), the answer to (c) is
that lim
x→0f(x) = −4.
(d) lim
x→2−f(x) = lim
x→2−¡x2−4¢= 22−4 = 0.
(e) lim
x→2+ f(x) = lim
x→2+ (4x−8) = 4 ·2−8 = 0. Since we got the same answer in (d), the answer to (f) is that
lim
x→2f(x) = 0. (g) We calculate the derivative of each piece of f(x): f0(x) =
2x+ 3 if x < 0
2xif 0 < x < 2
4 if x > 2.
The only values of xwhere f0(x) might not exist are x= 0 and x= 2. At x= 0 the slope coming from the
left is 2 ·0 + 3 = 3, and the slope coming from the right is 2 ·0 = 0, so f0(0) does not exist. At x= 2 the
slope coming from the left is 2 ·2 = 4, and so is the slope coming from the right, so f0(2) does exist, and it
equals 4. The graph of f(x) looks like
