Math 1A: Homework 2 Solutions
July 1
1. Prove the following claims:
(a) limx→a c = c for any real numbers a and c.
Let > 0 be given. Choose δ = 1. Suppose that 0 < |x − a| < δ. Then, we have
|f (x) − c| = |c − c| = 0 < so we conclude that
lim c = c.
x→a
(b) limx→2+
√
x − 2 = 0.
Let > 0 be given. √
Choose δ = 2 . √
Suppose that 0 < (x − 2) < δ. Then, we have
2
0 < (x − 2) < ⇒ x − 2 < ⇒ | x − 2 − 2| < . We conclude that
√
lim+ x − 2 = 0.
x→2
(c) limx→−1 (x2 − 5x − 2) = 4.
Let > 0 be given. Choose δ = min{1, 1000
}. Suppose that 0 < |x − (−1)| < δ.
Then, we have |x + 1| < 1000 . We also have |x + 1| < 1 ⇒ −1 < x + 1 < 1 ⇒
−8 < x − 6 < −6 ⇒ |x − 6| < 8. Thus,
|x + 1||x − 6| <
(8) < 1000
⇒ |x2 − 5x − 6| < ⇒ |(x2 − 5x − 2) − 4| < .
We conclude that limx→−1 (x2 − 5x − 2) = 4.
(d) limx→−1.5
9−4x2
3+2x
= 6.
Let > 0 be given and choose δ = 2 . Suppose that 0 < |x − (−1.5)| < δ. We
1
then have
⇒
⇒
⇒
⇒
⇒
⇒
We conclude that limx→−1.5
(e) limx→3
1
x
0 < |x + 1.5| < /2
0 < |2x + 3| < |2x + 3|
|2x + 3|
<
|2x + 3|
2
4x + 12x + 9 <
2x + 3
−4x2 − 12x − 9 <
2x + 3
9 − 4x2 − 6(2x + 3) <
2x + 3
9 − 4x2
2x + 3 − 6 < .
9−4x2
3+2x
= 6.
= 13 .
Let > 0 be given and choose δ = min{1, 1000
}. Suppose that 0 < |x − 3| < δ.
We then have |x − 3| < 1000 . We also have |x − 3| < 1 ⇒ −1 < x − 3 < 1 ⇒ 2 <
1
< 16 . Thus, we have
x < 4 ⇒ 6 < 3x < 12 ⇒ |3x|
1
|3x|
|x − 3|
3 − x
<
⇒ 3x 1 1
⇒ − < .
x 3
We conclude that limx→3
(f) limx→5
√
x=
√
1
x
1
<
<
1000
6
= 13 .
5. (Hint: rationalize!!)
Let > 0 be given and choose δ = min{1, 1000
}. Suppose that 0 < |x − 5|√< δ.
We then have |x − 5| < 1000 . We also have |x − 5| < 1 ⇒ 4 < x < 6 ⇒ 2 < x <
2
3⇒2<
√
x+
√
5<6⇒
1
√ √
| x+ 5|
< 12 . Thus, we have
1
1
√
<
<
|x − 5| √
1000
2
| x + 5|
x−5 √ <
⇒ √
x + 5
√
( x − √5)(√x + √5) √
⇒ <
√
x+ 5
√
√ ⇒ x − 5 < .
We conclude that limx→5
√
x=
√
5.
2. (a) Let f (x) = sin(x). By a graphical argument or otherwise, find an angle φ such
that f (x + φ) = cos(x).
1.5
1
0.5
0
-0.5
-1
-1.5
-6
-4
-2
0
2
4
6
8
10
Figure 1: sin(x) (solid) and cos(x) (dashed)
Observe that in Figure 1, we can get the graph of cos(x) by translating the graph
of sin(x) by π/2 units to the left. Thus, we have cos(x) = sin(x+π/2) = f (x+π/2)
so φ = π/2. In fact, we could have φ = π/2 + 2kπ for any integer k.
(b) Given that tan(π/2 − x) = cot(x), sketch the graph of g(x) = cot(x).
Let g(x) = tan(x). Then,the graph of h(x) = g(−x) = tan(−x) can be obtained
by reflecting the graph of g in the y-axis. Next, let f (x) = h(x − π/2) = g(π/2 −
x) = tan(π/2 − x) = cot(x). The graph of this can be obtained by moving the
graph of h by π/2 units to the right. We therefore end up with the graph shown
below in Figure 2.
3
20
10
0
-10
-20
-3
-2
-1
0
1
2
3
Figure 2: f (x) = cot(x)
3. Sketch the graphs of the following functions and use them to identify the points at
which limits do not exist.

2x + 1
x<0



cos(x) 0 ≤ x ≤ π
(a) f (x) =
sin(x) π < x ≤ 2π


 4x
e
x > 2π.
15
10
5
0
-5
-2
0
2
4
6
8
10
12
Figure 3: The e4x is not to scale
As can be seen in Figure 3, only at x = π, 2π the one-sided limits do not agree
with one another. Therefore, the limits do not exist at these points.
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

4 − x2
x<1
|x − 2|
1≤x≤5
(b) f (x) =

(x − 3)3 − 5
x > 5.
25
20
15
10
5
0
-5
-2
0
2
4
6
8
Figure 4
As can be seen in Figure 4, only at x = 1 the one-sided limits do not agree with
one another. Therefore, the limit does not exist at this point.
4. Sketch the following functions:
1
(a) f (x) = x+5
.
This is just 1/x translated 5 units to the left along the x-axis. See Figure 5.
(b) f (x) = 3 ln(x − 1).
Take ln(x), translate it one unit to the right along the x-axis and stretch it to
thrice its length along the y-axis. See Figure 6.
p
(c) f (x) = 16 − (x − 1)2 .
p
Write y = 16 − (x − 1)2 . Notice that y ≥ 0 and that
y 2 = 16 − (x − 1)2 ⇒ (x − 1)2 + y 2 = 16.
This is the equation of a circle of radius 4 centered at (1, 0). Since y ≥ 0, we only
get the upper semicircle. See Figure 7.
(d) f (x) = ex − e−x .
Draw ex and e−x and subtract the latter from the former. See Figure 8.
5. Find the largest possible domains for the following functions so they are one-one.
(a) f (x) =
ex +e−x
.
2
5
20
15
10
5
0
-5
-10
-15
-20
-8
-7
-6
-5
-4
-3
Figure 5: f (x) =
-2
-1
0
1
x+5
10
5
0
-5
-10
-15
0
2
4
6
8
Figure 6: f (x) = 3 ln(x − 1)
6
10
5
4
3
2
1
0
-1
-3
-2
-1
0
1
2
3
4
5
p
16 − (x − 1)2
Figure 7: f (x) =
20
15
10
5
0
-5
-10
-15
-20
-3
-2
-1
0
1
Figure 8: f (x) = ex − e−x
7
2
3
12
10
8
6
4
2
0
-3
-2
-1
0
Figure 9: f (x) =
1
2
3
ex +e−x
2
See Figure 9 for the graph of f . Observe that for f to pass the horizontal line
test, only one branch should survive. Therefore, we can take the required domain
to be (−∞, 0] OR [0, ∞).
(b) f (x) = 8 − (x + 3)2 .
See Figure 10 for the graph of f . Observe that for f to pass the horizontal line
test, only one branch should survive. Therefore, we can take the required domain
to be (−∞, −3] OR [−3, ∞).
10
5
0
-5
-10
-7
-6
-5
-4
-3
-2
-1
Figure 10: f (x) = 8 − (x + 3)2
(c) f (x) = cot(3x).
8
0
1
See Figure 11 for the graph of f . Observe that for f to pass the horizontal line
test, only one branch should survive. Therefore, we can take the required domain
to be (0, π/3) or any π/3-translated version of it.
10
5
0
-5
-10
-1
-0.5
0
0.5
1
1.5
2
Figure 11: f (x) = cot(3x)
6. Give an example of real valued functions f and g (and a corresponding real number a)
such that limx→a f (x)g(x) exists but limx→a f (x) and limx→a g(x) do not exist.
Let
f (x) =
−1 , x < 0
1 , x≥0
Also let g = f . Observe then that both f and g do not have a limit at zero since the
two one-sided limits do not agree. However,
1 , x<0
f (x)g(x) =
1 , x≥0
Since f (x)g(x) ≡ 1 for all x, we conclude that limx→0 = 1.
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