Written Homework 1 Solutions
Section 2.2
12. Given the function:


1 + sin(x)
x≤0



cos(x)
0
≤
x
≤π
f (x) = 


 sin(x)
x≥π
(a) Draw the graph of f (x)
Solution: Place the following into Wolfram Alpha.
plot Piecewise[1+sin(x),x¡0,cos(x),0¡=x¡=pi,sin(x),x¿pi]
Or see attached graph.
(b) Describe for which a the function has a limit.
Solution: First note that 1 + sin(x) is continuous on x ∈ (−∞, 0), cos(x) is continuous on x ∈ [0, π],
and sin(x) is continuous on x ∈ (π, ∞). Therefore the limit exists on all of those intervals except at the
interfaces. Those we still need to check. Now look to the first interface at 0.
lim 1 + sin(x) = 1
x→0−
lim cos(x) = 1
x→0+
Therefore one has
lim f (x) = 1
x→0
So then the limit as x approaches zero exists. Now to check the other interface.
lim cos(x) = −1
x→π−
lim sin(x) = 0
x→π+
So then:
lim f (x) = DNE
x→π
Then f (x) has a limit in the following domain: x ∈ (−∞, π) ∪ (π, ∞).
Homework 1 Solutions
Section 2.3
6. Evaluate the limit and justify each step by indicating the appropriate Limit Law. limu→2
Step
√
lim
u4 + 3u + 6
u→2
p
4
plimu→2 (u + 3u + 6)
2
plimu→2 (u ) + limu→2 (3u) + limu→2 (6)
4
Solution:
√(2) + limu→2 (3u) + limu→2 (6)
√16 + 3(2) + limu→2 (6)
√16 + 6 + 6
√28
2 7
Limit Law Number
•
11
1
9
3
7
•
•
26. Evaluate the limit of:
1
1
−
t t2 + t
!
1
1
lim − 2
t→0 t
t +t
!
lim
t→0
Solution:
1 t+1
1
− 2
lim ∗
t→0 t
t+1 t +t
!
t+1−1
lim
t→0
t2 + t
!
t
lim
t→0 t(t + 1)
!
1
lim
t→0 t + 1
Note that
1
t+1
!
is continuous at 0 so one can substitute for the limit. This means that:
!
!
1
1
1
1
lim − 2
= lim
=
=1
t→0 t
t→0 t + 1
0+1
t +t
52. Let f (x) = [[cos(x)]], on x ∈ [−π, π]
(a) Sketch the graph of f :
Solution: Place the following code into the search of Wolfram Alpha:
plot floor(cos(x))
Or see attached graph.
(b) Evaluate each limit if it exists.
i.
lim f (x)
x→0
Page 2 of 6
√
u4 + 3u + 6
Homework 1 Solutions
Solution: First check the left limit. Note that cos(x) is a value between 0 and 1 as x approaches 0.
The greatest integer for values between 0 and 1 is 0.
lim [[cos(x)]] = 0 lim+ [[cos(x)]] = 0 lim [[cos(x)]] = 0
x→0−
x→0
x→0
ii.
lim f (x)
x→( π2 )−
Solution: Look to the graph. From the left side one has:
lim [[cos(x)]] = 0
x→( π2 )−
iii.
lim f (x)
x→( π2 )+
Solution: Look to the graph. From the right side one has:
lim [[cos(x)]] = −1
x→( π2 )+
iv.
lim f (x)
x→( π2 )
Solution: In the previous two parts one found the limit from each side. Since they do not agree
one has:
limπ [[cos(x)]] = DNE
x→( 2 )
(c) Where does the limit exist?
Solution: Looking at the graph, one has a limit from the left and the right for all values of x except at
± π2 . Then one has a limit for x ∈ (−π, − π2 ) ∪ (− π2 , π2 ) ∪ ( π2 , π).
Page 3 of 6
Homework 1 Solutions
Section 2.5
18. Explain why the function is discontinuous at the given number a. Sketch the graph of the function.
( 1
x , −2
f (x) = x+2
1
x = −2
Where we check the continuity at a = −2.
Solution: For the graph one can look at the attached graphs.
Looking at the graph one can see clearly that this function is discontinuous. Check the continuity conditions
anyway.
First note that f (a) = 1 and does exist.
Then one needs to check the limit at a. Check the left and right limits.
lim f (x) = −∞
x→−2−
lim f (x) = ∞
x→−2+
Therefore one has that:
lim f (x) = DNE
x→−2
The function is discontinuous there.
46. Find the values of a and b that make f continuous everywhere.
 x2 −4

x<2



 x−22
ax − bx + 3 2 ≤ x < 3
f (x) = 



 2x − a − b
3≤x
Solution: In order to be continuous the function must have a limit equal to the value at each point. Since
each function on its interval is continuous one only need to check the interfaces.
lim− f (x) = lim−
x→2
x→2
x2 − 4
(x + 2)(x − 2)
= lim−
= lim− x + 2 = 4
x→2
x→2
x−2
x−2
So then ax2 − bx + 3 = 4 at x=2 for continuity at that interface. At the same time ax2 − bx + 3 = 2x − a + b
at x=3 for continuity there. First solve for a in terms of b in the x=3 equation.
a(3)2 − b(3) + 3 = 2(3) − a + b
10a − 4b = 3
3 + 4b
a=
10
Now to place this equation for a into the x=2 equation.
a(2)2 − b(2) + 3 = 4
4a − 2b = 1
3 + 4b
4
− 2b = 1
10
6 + 8b 10b
−
=1
5
5
Page 4 of 6
Homework 1 Solutions
6 − 2b
=1
5
6 − 2b = 5
−2b = −1
1
b=
2
Then recover a with the following.
a=
5
1
3 + 4b 3 + 2
=
=
=
10
10
10 2
So then at each interface the limits exist, have a value, and the limit equals the value. The function is now
continuous everywhere.
Page 5 of 6
Homework 1 Solutions
50. Suppose f is continuous on [1, 5] and the only solutions of the equation f (x) = 6 are x = 1 and x = 4. If
f (2) = 8, explain why f (3) > 6.
Solution: Perform a proof by contradiction. Assume that α ∈ (2, 4) such that f (α) ≤ 6. Then by the
Intermediate Value Theorem (letting a = 2, b = α, and N = 6) there exists a c such that f (c) = 6. However
this violates the definition of f (x) since c , 1 and c , 4. Therefore the initial assumption that α is such that
f (α) < 6 is false. Therefore f (α) > 6. Since 3 is a viable choice for α, f (3) > 6.
Page 6 of 6