MATHEMATICS 100, Section 103
Midterm #1, October 3, 2007
x2 − 4
(x − 2)(x + 2)
x+2
4
= lim
= lim
= .
2
x→2 x − x − 2
x→2 (x + 1)(x − 2)
x→2 x + 1
3
1. a) lim
b) lim
x→∞
p
√
√
2 + 4x − x)( x2 + 4x + x
x
(
√
x2 + 4x − x = lim
x→∞
x2 + 4x + x
x2 + 4x − x2
4
√
= lim
= lim p
= 2.
2
x→∞
x→∞
x + 4x + x
1 + 4/x + 1
c) lim
x→3+
|x − 3|
x−3
= lim
= 1.
+
|x| − 3 x→3 x − 3
d) Let f (x) = x2 − cos x, then f is continuous for all x. We have f (0) = 0 − 1 =
−1 < 0 and f (π) = π 2 + 1 > 0. By the Intermediate Value Theorem, there is an
x ∈ (0, π) such that f (x) = 0. This x solves the equation x2 = cos x.
e) Since 1 is a root of x2 − 3x + 2 = 0, the limit can only exist if the numerator is 0
at x = 1, i.e. 12 + 1 + a + a + 4 = 0, 2a + 6 = 0, a = −3. For this value of a, we
have
x2 − 2x + 1
(x − 1)2
x2 + x + ax + a + 4
=
lim
=
lim
x→1 x2 − 3x + 2
x→1 (x − 1)(x − 2)
x→1
x2 − 3x + 2
lim
x−1
= 0.
x→1 x − 2
= lim
f) We have f (x) = e2x + ex , hence f 0 (x) = 2e2x + ex .
1
g) f (x) =
2
0
h) If g(x) = x2
sin x
x + cos(2x)
−1/2
·
cos x(x + cos(2x)) − sin x(1 − 2 sin(2x))
.
(x + cos(2x))2
p
f (x) then
p
√
1
1
g 0 (x) = 2x f (x) + x2 · p
· f 0 (x), g 0 (2) = 4 4 + 4 · √ · 3 = 8 + 3 = 11.
2 4
2 f (x)
2. Let y(t) = t2 − 3t + 6. The average velocity between t = 1 and t = 3 is
y(3) − y(1)
(9 − 9 + 6) − (1 − 3 + 6)
6−4
=
=
= 1m/sec.
3−1
2
2
We have y 0 (t) = 2t − 3, hence the instantaneous velocity at t = 2 is y 0 (2) = 4 − 3 =
1m/sec.
3. We have
lim g(x) = lim (x2 + ax + b) = 1 + a + b,
x→1−
x→1−
lim g(x) = lim x3 = 1.
x→1+
x→1+
For g(x) to be continuous at 1, the two limits need to be equal, i.e. 1 + a + b = 1,
a + b = 0. For g(x) to be differentiable at 1, it should be continuous at 1 and, in
addition,
d 2
d 3
(x + ax + b)|x=1 =
x |x=1 , 2x + a|x=1 = 3x2 |x=1 , 2 + a = 3,
dx
dx
hence a = 1 and b = −1, where we also used the condition a + b = 0 for continuity.
4. Let (a, b) be the tangency point, then the slope of the curve at (a, b) is 7, i.e. y 0 (a) =
3a2 − 5 = 7, 3a2 = 12, a2 = 4, a = ±2.
If a = 2, then b = 7 · 2 − 1 = 13, and we can find m from
23 − 5 · 2 + m = 13, 8 − 10 + m = 13, m = 15.
If a = −2, then b = 7 · (−2) − 1 = −15, and we can find m from
(−2)3 − 5 · (−2) + m = 13, −8 + 10 + m = −15, m = −17.
√
√
x+h−3− x−3
h
√
√
√
√
( x + h − 3 − x − 3)( x + h − 3 + x − 3)
√
√
= lim
h→0
h( x + h − 3 + x − 3)
d √
5.
x − 3 = lim
h→0
dx
x+h−3−x+3
1
1
√
√
√
= lim √
= √
.
h→0 h( x + h − 3 +
2 x−3
x − 3) h→0 x + h − 3 + x − 3
= lim