Math 120 (Winter 2015-16) HW #3, Due on Wednesday, Sept. 30.
1. Show, using a δ − proof, that limx→3 x3 = 27.
|x3 − 27| = |x − 3|||x2 + 3x + 9|
If |x − 3| < 1, then |x| < 4 and so
|x2 + 3x + 9| ≤ |x|2 + 3|x| + 9 < 37
Let > 0 and δ := min(/37, 1). If |x − 3| < δ, then
|x3 − 27| = |x − 3||x2 + 3x + 9| < (/37)(37) = .
2. Let
f (x) =
0 x rational
x x irrational
Recall that in class that we showed, using the formal definition of limit, that limx→0 f (x) =
0. Prove this using the squeeze theorem instead.
Let
x x≥0
0 x<0
0 x≥0
x x<0
g(x) =
h(x) =
Then h(x) ≤ f (x) ≤ g(x) and limx→0 g(x) = 0 = limx→0 h(x).
By the squeeze theorem, limx→0 f (x) = 0.
4
−100000x+99999
√
3. Find limx→∞ 3x
. Explain your answer but you do not need to justify your
2x4 +37912x2 + 2x
answer with a formal definition of limit.
Dividing numerator and denominator by x4 , we have
3x4 − 100000x + 99999
3 − 100000x−3 + 99999x−4
√
√
= lim
lim
x→∞
x→∞ 2x4 + 37912x2 +
2x
2 + 37912x−2 + 2x−3
= 3/2.
1
1
4. Find limx→∞ √x2 +5x−x
and limx→−∞ √x2 +5x−x
. Explain your answers but you do not
need to justify your answers with a formal definition of limit.
√
1
x2 + 5x + x
√
√
= √
x2 + 5x − x
( x2 + 5x − x)( x2 + 5x + x)
√
√
x2 + 5x + x
x2 + 5x + x
= 2
=
x + 5x − x2
5x
1
Dividing numerator and denominator by x, we obtain (for x > 0) that this equals
q
p
x2 +5x
+1
1 + 5/x + 1
x2
=
.
5
5
So,
p
1 + 5/x + 1
= 2/5.
5
√
As x → −∞, x + 5 → −∞ and so x(x + 5) → ∞ and so x2 + 5x → ∞
√
As x → −∞, −x → ∞ and so x2 + 5x − x → ∞.
1
= lim
lim √
x→∞
x2 + 5x − x x→∞
Thus, limx→−∞
√
1
x2 +5x−x
= 0.
5. Using the formal definition of limit at ∞ (Definition 10, page 91) show that
1
limx→∞ √x−1
= 0.
Let > 0. We need to find R > 0 s.t. if x > R, then
1
1
|√
− 0| = | √
| < .
x−1
x−1
1
Assuming x > 1, we have | √x−1
|=
√1 .
x−1
1
< . Solving for an equivalent inequality on x, we get
So, we want √x−1
√
x − 1 > 1/ and so
√
x > 1/ + 1 and so
x > (1/ + 1)2 .
These steps are all reversible, so if we let R := (1/ + 1)2 and x > R, then
And if x > R, then x > 1 and we get
|√
√1
x−1
< .
1
1
|= √
< .
x−1
x−1
6. For what value(s) of c is the function
2x − c x ≤ 3
f (x) =
5 − cx x > 3
continuous on R (i.e., continuous at all a ∈ R).?
Since polynomials are continuous everywhere, f (x) is continuous at all x 6= 3.
By the limit theorems for polynomials, limx→3 2x−c = 6−c and limx→3 5−cx = 5−3c.
So, limx→3− f (x) = 6 − c and limx→3+ f (x) = 5 − 3c.
So, f is continuous at 3 (and hence everywhere) if and only if 6−c = 5−3c, equivalently
c = −1/2.
2
√
√
√
4
7. Let f (x) = xx2 −9
3}.
Define
f
(
3)
and
f
(−
3) s.t. f is continuous on R
on
{x
=
6
±
−3
(i.e., continuous at all a ∈ R).
√
For x 6= ± 3,
f (x) = x2 + 3
√
Thus for a = √
± 3, limx→a
√ f (x) = 6. So, f extends to a continuous function on R if
and only if f ( 3) = f (− 3) = 6.
√
8. Show, using a δ − proof, that f (x) = 4 x is continuous on [0, ∞) (i.e., continuous at
all a ∈ [0, ∞)).
Case 1: a > 0.
√
4
|f (x) − f (a)| = | x −
√
4
√
√
| x − a|
√
a| = √
4
x+ 4a
|x − a|
|x − a|
√
√
√ ≤ √
√
= √
4
4
4
( x + a)( x + a)
a a
Let > 0. We want |f (x) − f (a)| < . By the above, this will hold if
|x − a|
√ < .
√
4
a a
Note that
√
√
4
a a = a3/4 . Let δ := a3/4 . If |x − a| < δ, then we have
a3/4 |x − a|
√
√
√ < .
<
|f (x) − f (a)| ≤ √
4
4
a a
a a
Case 2: a = 0.
√
√
√
4
4
4
4
4
x
−
0|
=
x<
Let
>
0
and
δ
:=
.
If
|x
−
a|
<
δ,
then
x
<
δ
=
,
and
so
|
√
4 4
= .
3