MATH 100, Section 110 (CSP)
Week 1: Marked Homework Solutions
2010 Sep 16
1. (a) limt→−2
t2 −4
t+2
= limt→−2
(t+2)(t−2)
t+2
= limt→−2 (t − 2) = −4.
2
−3x+2
(b) limx→1 xx2 −2x+1
= limx→1 (x−1)(x−2)
does not exist. If x is near 1 and x > 1
= limx→1 x−2
(x−1)2
x−1
(for example, x = 1.001), then x − 2 is near −1 and x − 1 is near 0 and positive, therefore
x−2
≈ −1
is a negative number with a large absolute value. On the other hand, if x is near
x−1
+0
1 and x < 1 (for example, x = 0.999), then x − 2 is still near −1 but x − 1 is near 0 and
x−2
negative, therefore x−1
≈ −1
is a positive number with a large absolute value. So x−2
cannot
−0
x−1
be made close to a unique finite number L for all x near 1 (both x > 1 and x < 1). (It is
2
2
also true that limx→1+ xx2 −3x+2
= −∞ and limx→1− xx2 −3x+2
= ∞, but to answer the given
−2x+1
−2x+1
question correctly you must state that the limit does not exist.)
(c) limx→−1
x−1
x2 −2x+1
= limx→−1
x−1
(x−1)2
= limx→−1
1
x−1
=
−1
−1−1
= − 12 .
x−2
x−2
1
(d) limx→2 x2 −4x+4
= limx→2 (x−2)
2 = limx→2 x−2 does not exist, for reasons similar to those
in (b). If x is near 2 and x > 2 (for example, x = 2.00001), then x − 2 is near 0 and positive,
1
1
≈ +0
is a positive number with a large absolute value. On the other hand, if x
therefore x−2
is near 2 and x < 2 (for example, x = 1.99999), then x − 2 is near 0 and negative, therefore
1
1
1
≈ −0
is a negative number with a large absolute value. So x−2
cannot be made close
x−2
to a unique finite number L for all x near 2 (both x > 2 and x < 2). (While it is true that
x−2
x−2
= ∞ and limx→2− x2 −4x+4
= −∞, to answer the given question correctly
limx→2+ x2 −4x+4
you must state that the limit does not exist.)
(e) limh→0
(f) limh→0
limh→0
1
h
·
1
h
1
a+h
−
1
a
= limh→0
1 a−(a+h)
h (a+h)a
= limh→0
−1
(a+h)a
=
−1
a·a
= − a12 .
√ √
√ √
√ √
x− x+h
x− x+h
x+ x+h
1
1
√1
√
√
√1
√
√
√ √
−
=
lim
·
=
lim
·
·
=
h→0 h
h→0 h
x
x+h
x+h x
x+h x
x+ x+h
x−(x+h)
√
√ √
√ √ √
√ √
= limh→0 √x+h√x (−1
= √x√x (−1
= − 2x1√x .
x+ x)
x+h x( x+ x+h)
x+ x+h)
1
h
2. First think of the radius r > 0 of the circle C2 as fixed, and:
• Find the coordinates of Q.
• Find the slope, and then the equation, of the line P Q.
• Find the coordinates of R (f (r), 0), i.e. the x-intercept f (r) of the line P Q.
Then think of r as a variable and evaluate limr→0+ f (r).
The equation for C2 is x2 + y 2 = r2 , or y 2 = r2 − x2 . Similarly, the equation for C1 can
be written y 2 = 4 − (x − 2)2 , therefore Q corresponds to a solution of r2 − x2 = 4 − (x − 2)2 .
Solving for xrgives the x-coordinate of Q as x = r2 /4, then the y-coordinate of Q is y =
q
2 2
√
r2
r2 − x2 = r2 − r4 = r 1 − 16
(taking the positive square root to get the intersection
point in the upper half-plane).
1
q
The slope of the line passing through P (0, r) and Q
4
r
q
1−
r2
16
r2
,r
4
q
1−
r2
16
is mP Q =
2
1− r16 − r
r
r2
−0
4
=
− 1 , and the equation of the line is y = mP Q x + r.
The x-coordinate of R is the x-intercept of the line y = mP Q x + r, so set y = 0 in the
r2
. This last expression
equation for the line and solve for x to get x = − mPr Q = q
2
1− r16
4 1−
is f (r).
Then as r shrinks to 0 we have
q
r2
limr→0+ f (r) = limr→0+ q
lim
r→0+
4 1+
q
1−
r2
16
2
1− r16
4 1−
r2
= limr→0+ q
4 1−
2
·
1− r16
1+
q
1+
r2
2
1− r16
2
1− r16
= limr→0+
q
1+
h
2
1− r16
2
4 1−(1− r16 )
i
= 8.
R approaches the point (8, 0) as C2 shrinks to the origin.
3. (a) If x → 0− then x is strictly negative and we use the definition of f (x) = Ax2 + Bx + C
for x negative. The function Ax2 + Bx + C is a polynomial, so we can use the Direct
Substitution Property (p. 102, and see remark just below Theorem 1 on p. 104) to compute
lim f (x) = lim (Ax2 + Bx + C) = C.
x→0−
x→0−
(b) If x → 0+ then x is strictly positive and we use the definition of f (x) = x3/2 cos(1/x)
for x positive. The function 1/x is not defined at x = 0, and we cannot merely substitute
x = 0 to find limx→0+ f (x).Instead, we use the Squeeze Theorem (p. 105): since it is
always true that
−1 ≤ cos(1/x) ≤ 1
for all x near 0, this implies that
−x3/2 ≤ x3/2 cos(1/x) ≤ x3/2
for all positive x near 0 (what if x was negative?); then since both
√ lim+ −x3/2 = lim+ −x x = 0
x→0
and
x→0
√
lim+ x3/2 = lim+ x x = 0,
x→0
x→0
(using the Root Law, p. 101) we have by the Squeeze Theorem
lim f (x) = lim+ x3/2 cos(1/x) = 0.
x→0+
x→0
(c) In the first line of the definition of the function f , it says f (x) = Ax2 + Bx + C if
−∞ < x ≤ 0. Since x = 0 is included in this part of the definition, we just plug x = 0 into
the formula Ax2 + Bx + C and get
f (0) = C
by the definition of the function.
2
=