Solutions to Graded Problems Math 200
Section 2.3
38. Prove that limx→0+
Solution.
√
Homework 3
September 24, 2010
x esin(π/x) = 0.
Note that −1 ≤ sin(π/x) ≤ 1 for all x ∈ R \ {0}. In particular then,
√ −1 √ sin(π/x) √ 1
xe ≤ xe
≤ xe
√
√
√
x
≤ lim+ x esin(π/x) ≤ lim+ xe1
lim+
x→0
x→0
x→0
e
0 ≤ lim+
√
x esin(π/x) ≤ 0.
x→0
Thus, limx→0+
√
xesin(π/x) = 0 by the Squeeze Theorem.
50. Let f (x) = bcos(x)c, −π ≤ x ≤ π. (This is the floor function and it is equivalent to
the greatest integer function in your book.)
(a) Sketch the graph of f .
Solution.
y
1
−4π
4
−3π
4
−2π
4
−1π
4
1π
4
2π
4
3π
4
4π
4
x
-1
(b) Evaluate each limit, if it exists:
(i) lim f (x)
x→0
(iii)
lim
x→(π/2)+
f (x)
(ii)
lim
x→(π/2)−
f (x)
(iv) lim f (x)
x→π/2
Solution. (i) Note that as x → 0 from both sides, the function maintains a value of zero,
so limx→0 f (x) = 0 (even though f (0) = 1.
(ii) By inspection, we see that limx→(π/2)− f (x) = 0.
(iii) Note, limx→(π/2)+ f (x) = −1.
(iv) Since our answers in parts (ii) and (iii) are different, this limit does not exist: limx→(π/2)+ f (x)
DNE.
(c) For what values of a does limx→a f (x) exist?
Solution. The only places where we have trouble is ± π2 where the right and left handed
limits do not agree. Also, at ±π one can only take a right handed or left handed limit. So,
f doesn’t have true limits here either.
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Solutions to Graded Problems Math 200
Homework 3
September 24, 2010
Section 2.5
30. Locate the discontinuities of the function y = ln(tan2 (x)) and illustrate by graphing.
Solution. Note that since ln(x) and tan2 (x) are continuous everywhere in their domains,
y = ln(tan2 (x)) is as well by composition. Thus, it suffices to find the domain of y. Note
that cos(π/2 + kπ) = 0 for all integers k ∈ Z, those points are out. Next, note that at kπ
. A graph
where k ∈ Z, we have that tan(x) = 0. This rules out all numbers of the form kπ
2
is sketched below:
38. Find the numbers at which f is discontinuous. At which of these numbers is f continuous from the right, form the left, or neither? Sketch the graph of f .

x≤1
 x + 1 if
1/x
if 1 < x < 3
f (x) =
 √
x − 3 if
x≥3
Solution. Since each of these functions is continuous everywhere on its domain, it suffices to check only at the points where we transition from one function to another. That is,
we only need check 1 and 3. Note that
lim x + 1 = 2 6= 1 = lim
x→1
x→1
1
x
so f is not continuous at 1. In a similar fashion,
√
1
= 1 6= 0 = lim x − 3
x→1 x
x→3
lim
so f is not continuous at 3. Here is a graph:
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Solutions to Graded Problems Math 200
Homework 3
September 24, 2010
y
3
2
1
-1
1
2
3
x
-1
Section 2.7
x2 + 1
.
28. Find f (a) if f (x) =
x−2
0
Solution. We use the definition of the derivative f 0 (a) = limh→0
f 0 (a). We have
0
f (a) = lim
h→0
(a2 +h)2 +1
(a+h)−2
−
f (a+h)−f (a)
h
to compute
a2 +1
a−2
h
(a + h)2 + 1
a2 + 1
−
h→0 h(a + h − 2)
h(a − 2)
= lim
(a − 2)((a + h)2 + 1) − (a + h − 2)(a2 + 1)
h→0
h(a + h − 2)(a − 2)
= lim
(a − 2)(a2 + 2ah + h2 + 1) − (a3 + ha2 − 2a2 + a + h − 2)
h→0
h(a + h − 2)(a − 2)
= lim
a3 + 2a2 h + ah2 + a − 2a2 − 4ah − 2h2 − 2 − a3 − ha2 + 2a2 − a − h + 2
h→0
h(a + h − 2)(a − 2)
= lim
a3 (1 − 1) + a2 (2h − 2 − h + 2) + a(h2 + 1 − 4h − 1) − h(2h + 1)
h→0
h(a + h − 2)(a − 2)
= lim
ha2 + ah2 − 4ah − h(2h + 1)
= lim
h→0
h(a + h − 2)(a − 2)
h(a2 + ah − 4a − 2h − 1)
h→0
h(a + h − 2)(a − 2)
= lim
3
Solutions to Graded Problems Math 200
Homework 3
September 24, 2010
a2 + ah − 4a − 2h − 1
h→0
(a + h − 2)(a − 2)
= lim
=
a2 − 4a − 1
(a − 2)2
=
a(a − 4) − 1
.
(a − 2)2
Note that on the sixth line from the bottom, the terms are grouped according to their power
of a and then h. You don’t have to do it this way yourself, but you may find it easier to
group by like terms to make sure you don’t miss anything.
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