Download Math 130A Homework Set 8 Partial Solutions

Math 130A Homework Set 8 Partial Solutions
1. Let f (x) = 3x2 − 2x + 1. Show that f is continuous at 2.
Proof. Let > 0. Consider
|f (x) − f (2)| = |3x2 − 2x + 1 − 9| = |x − 2||3x + 4|.
As x → 2, so assume that |x − 2| < 1, then 1 < x < 3 =⇒ 3x + 4 < 13. Hence choose
δ = min{1, /13}, such that for all |x − 2| < δ =⇒ |f (x) − f (2)| < .
2. Let f : [−4, 0] → R defined as follows
f (x) =
2x2 −18
x+3
: x 6= −3
−12 : x = −3.
Show that f is continuous at −3.
Proof. Let > 0. Then |f (x) − f (−3)| = |f (x) + 12| = |2x − 6 + 12| = 2|x + 3|. So we
choose δ = /2, such that |x + 3|δ =⇒ |f (x) − f (−3)| < .
4. If x0 ∈ E, x0 is not an accumulation point of E, and f : E → R. Prove that for every
sequence {xn } with xn → x0 with xn ∈ E for all n, f )xn ) → f (x0 ).
Proof. (Sketch of proof) As x0 is not an accumulation point of E, so for xn → x0 with
xn ∈ E implies that {xn } must be eventually constant, i.e., there exists N such that
xn = x0 for n ≥ N . Thus f (xn ) = f (x0 ) for n ≥ N , so f (xn ) → f (x0 ) .
6. Prove that
√
x is continuous for all x ≥ 0.
Proof. Let x0 ≥ 0. We want show f (x) is continuous at x0 . Suppose that x0 = 0.
Then f (0) = 0. So for > 0, choose δ = 2 , such that |x − 0| < δ and x ≥ 0 implies
that
√
√
√
| x − 0| = x < 2 = .
So f is continuous at 0.
Now x0 > 0. Then
x − x |x − x |
√
√
0 0
|f (x) − f (x0 )| = | x − x0 | = √
.
√ ≤ √
x0
x + x0
√
Hence we choose δ = x0 such that |x − x) | < δ =⇒ |f (x) − f (x) | < . Hence f is
continuous at x0 .
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8. Suppose that f : (a, b) → R isa continuous and f (r) = 0 for each rational number
r ∈ (a, b). Prove that f (x) = 0 for all x ∈ (a, b).
Proof. (Sketch of proof) Let x ∈ (a, b). If x is rational, then f (x) = 0. Now let x
is irrational number. Then from the density of rational numbers, there is a sequence
{rn } such that rn → x, where rn is rational for each n. Then apply f is continuous at
x, limn→∞ f (rn ) = f (x), but f (rn ) = 0 for each n, hence f (x) = 0.
13. Let f : D → R be continuous at x0 ∈ D. Prove that ∃M > 0 and a neighborhood Q
of x0 such that |f (x)| ≤ M for all x ∈ Q ∩ D.
Proof. As f is continuous at x0 , by definition, ∀ > 0∃δ > 0 such that ∀x ∈ D with
|x − x0 | < δ =⇒ |f (x) − f (x0 )| < . In particular, let = 1, there exists a δ > 0 such
that for all x ∈ D with |x − x) | < δ, we have |f (x) − f (x0 )| < 1. Thus,
|f (x)| ≤ |f (x) − f (x0 )| + |f (x0 )| < 1 + |f (x0 )|, for all x ∈ D, |x − x0 | < δ.
14. If f : D → R is continuous at x0 ∈ D, prove that |f | : D → R with |f |(x) = |f (x)| is
continuous at x0 .
Proof. (Sketch of proof) As f is continuous at x0 , so ∀ > 0∃δ > 0 such that x ∈ D
and |x − x0 | < =⇒ |f (x) − f (x0 )| < . Note that ||f (x)| − |f (x0 )|| ≤ |f (x) − f (x0 )|.
So the above also says that ∀ > 0∃δ > 0 such that x ∈ D and |x − x0 | < =⇒
||f (x)| − |f (x0 )|| < . Hence |f | is continuous at x0 .
15. Suppose that f, g are continuous at x0 . Prove that h(x) = max{f (x), g(x)} is continuous at x0 .
Proof. (Sketch of Proof) Notice that
h(x) = max{f (x), g(x)} =
f (x) + g(x) + |f (x) − g(x)|
,
2
as we know f + g, f − g are continuous at x0 by Theorem 3.2, |f − g| is continuous by
the result of exercise 13.
17. Suppose that f ≥ 0 for all x ∈ D. Show that if f is continuous at x0 , then
continuous at x0 .
√
f is
√
Proof. (Sketch of Proof) As we know g(x) = x is continuous at all x ≥ 0, in p
particular, at f (x0 ) ≥ 0, and f is continuous at x0 , then by Theorem 3.4, g(f (x)) = f (x)
is continuous at x0 .
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1. Suppose that the function f is defined only on the integers. Explain why it is continuous. Hint. Each integer is not an accumulation point of integers.
2. Suppose that a function f is continuous at a point c and f (c) > 0. Prove that there is
f (c)
> 0.
a δ > 0 so that for all x ∈ Domain(f ), |x − c| < δ implies f (x) ≥
2
Proof. (Sketch of proof) As f is continuous at c, and f (c) > 0, let = f (c)
. Then there
2
f (c)
exists a δ > 0 such that |x − x0 | =⇒ |f (x) − f (c)| < . So f (c) − = 2 < f (x), x ∈
{x : |x − x0 | < δ}.
3. Let f (x) = x2 and let > 0 be given.
(a) Find a δ so that |x − 1| < δ implies |f (x) − 1| < .
(b) Find a δ so that |x − 2| < δ implies |f (x) − 4| < .
(c) If n > 2 and find a δ so that |x − n| < δ implies |f (x) − n2 |. Would the δ be
larger or smaller that the δ for parts (a) and (b)? Why?
Answer. (a). Choose δ = min{1, /3}.
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