Math 2250 Written HW #3 Solutions

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Math 2250 Written HW #3 Solutions
1. For which values of d is the function
(
d2 x − 2d
h(x) =
12
if x ≥ 2
if x < 2
continuous at every number x?
Answer: Remember that h(x) is continuous at x = 2 if and only if
lim h(x) = h(2).
x→2
Now, we know that h(2) = 2d2 − 2d, so that’s the right hand side of the above equation. As
for the left hand side, in order for limx→2 h(x) to exist, we need to have
lim h(x) = lim h(x).
x→2−
x→2+
But clearly
lim h(x) = lim 12 = 12.
x→2−
x→2−
On the other hand,
lim h(x) = lim
x→2+
x→2+
d2 x − 2d = 2d2 − 2d.
Therefore, we only have limx→2− h(x) = limx→2+ h(x) if
12 = 2d2 − 2d
or, equivalently,
0 = 2d2 − 2d − 12 = 2(d2 − d − 6) = 2(d − 3)(d + 2),
which only happens when d = 3 or d = −2.
Therefore, we only have limx→2− h(x) = limx→2+ h(x) when d = 3 or d = −2. In this case,
lim h(x) = lim h(x) = lim h(x) = 12.
x→2
x→2−
x→2+
But of course if d is either 3 or −2 then we know h(2) = 12 = limx→2 h(x), so we can conclude
that the function h(x) is continuous when d = 3 or d = −2.
Here are the graphs of h(x) when d = 3 and when d = −2:
30
d = -2
25
20
d=3
15
10
5
1
2
1
3
4
2. Give an example of two functions f (x) and g(x) so that both f (x) and g(x) are discontinuous
at x = 0, but the product f (x) · g(x) is actually continuous at x = 0.
Answer: Suppose
and
(
−1
f (x) =
1
if x ≤ 0
if x > 0
(
1
g(x) =
−1
if x ≤ 0
if x > 0.
Then
h(x) = f (x)g(x) = −1
for all x. Clearly, then, h(x) is continuous at x = 0 even though neither f nor g is continuous
at zero.
3. Evaluate the limit
3x3 − 12x2 − 1
.
x→∞ 9x3 + 2x + 13
lim
Answer: Notice that both the numerator and the denominator are going to infinity as
x → ∞, so we need to do some thinking. In this case, we have a polynomial in both the
numerator and the denominator, so the right thing to do is divide both by the highest power
of x that we can see, which is x3 :
3x3 − 12x2 − 1
3x3 − 12x2 − 1 1/x3
=
lim
·
x→∞ 9x3 + 2x + 13
x→∞ 9x3 + 2x + 13
1/x3
1
3 − 12x2 − 1
3x
3
= lim x 1
x→∞
(9x3 + 2x + 13)
x3
lim
= lim
x→∞
3
9
1
= .
3
=
2
3−
9+
12
x
2
x2
−
+
1
x3
13
x3
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