MATHEMATICS 100, Section 105
Midterm #1, October 3, 2007
x2 − 4x + 3
(x − 1)(x − 3)
x−3
−2
1
=
lim
=
lim
=
=
−
.
x→1
x→1 2(x − 1)(x + 1)
x→1 2(x + 1)
2x2 − 2
4
2
1. a) lim
b)
x3 + 2x
1 + 2x−2
1
1
=
lim
=
=− .
x→−∞ 1 − x2 − 4x3
x→−∞ x−3 − x−1 − 4
−4
4
lim
c) lim
x→2−
|x| − 2
x−2
= lim
= −1.
|x − 2| x→2− −(x − 2)
d) Let f (x) = x2 − e−x , then f (x) is continuous for all x. We have f (0) = 0 − e0 =
−1 < 0 and f (1) = 1 − e−1 ≈ 1 − (2.7)−1 > 0. By the Intermediate Value
Theorem, there is an x ∈ (0, 1) such that f (x) = 0. This x solves the equation
x2 = e−x .
e) Since 2 is a root of x2 − 5x + 6 = 0, the limit can only exist if the numerator is
0, i.e. 4 + 4 + 2a + a + 1 = 0, 3a + 9 = 0, a = −3. For this value of a, we have
x2 + 2x + ax + a + 1
x2 − x − 2
(x − 2)(x + 1)
=
lim
= lim
2
2
x→2
x→2 x − 5x + 6
x→2 (x − 2)(x − 3)
x − 5x + 6
lim
3
x+1
=
= −3.
x→2 x − 3
−1
= lim
f) We have f (x) = e−x + ex , hence f 0 (x) = −e−x + ex .
1
g) f (x) =
2
0
h) g 0 (x) =
p
cos(2x)
x − sin x
−1/2
·
−2 sin(2x) · (x − sin x) + cos(2x) · (1 − cos x)
.
(x − sin x)2
p
xf 0 (2x)
1
· f 0 (2x) · 2 = f (2x) + p
,
f (2x) + x · p
2 f (2x)
f (2x)
g 0 (1) =
√
f 0 (2)
−3
3
1
f (2) + p
= 4+ √ =2− = .
2
2
4
f (2)
p
2. Let y(t) = t2 − 2t + 6. The average velocity between t = 1 and t = 4 is
y(4) − y(1)
(16 − 8 + 6) − (1 − 2 + 6)
14 − 5
=
=
= 3m/sec.
4−1
3
3
We have t0 (t) = 2t − 2, hence the instantaneous velocity of the particle at t = 3 is
y 0 (3) = 6 − 2 = 4m/sec.
3. We have
lim g(x) = lim (x3 + ax + b) = 1 + a + b,
x→1−
x→1−
lim g(x) = lim 2x2 = 2.
x→1+
x→1+
For g(x) to be continuous at 1, the two limits need to be equal, i.e. 1 + a + b = 2,
a + b = 1. For g(x) to be differentiable at 1, it should be continuous at 1 and, in
addition,
d 3
d 2
(x + ax + b)|x=1 =
2x |x=1 , 3x2 + a|x=1 = 4x|x=1 , 3 + a = 4,
dx
dx
hence a = 1 and b = 0, where we also used the condition a + b = 1 for continuity.
4. Let (a, b) be the tangency point, then the slope of the curve at (a, b) is 5, i.e. y 0 (a) =
3a2 + 2 = 5, 3a2 = 3, a2 = 1, a = ±1.
If a = 1, then b = 5 − 1 = 4, and we can find m from
1 + 2 + m = 4, m = 1.
If a = −1, then b = −5 − 1 = −6, and we can find m from
−1 − 2 + m = −6, m = −3.
√
√
d
1
1
1
1 x−2− x+h−2
√
√
√
= lim
−√
= lim √
5.
h→0 h x + h − 2 x − 2
dx x − 2 h→0 h
x−2
x+h−2
√
√
√
√
( x − 2 − x + h − 2)( x − 2 + x + h − 2)
√
√
√
√
= lim
h→0 h x + h − 2 x − 2( x − 2 +
x + h − 2)
(x − 2) − (x + h − 2)
√
√
√
h→0 h x + h − 2 x − 2( x − 2 +
x + h − 2)
= lim
= lim √
h→0
√
1
1
√
√
√
√
=
.
2(x − 2) x − 2
x + h − 2 x − 2( x − 2 + x + h − 2)