SOLUTION OF HW7
MINGFENG ZHAO
November 06, 2011
1. [Page 142, Problem 19] Calculate the limit:
√
lim
x→0
1+x−
x
√
1−x
.
Proof. For any 0 < |x| < 1, we know that 1 + x > 0, and 1 − x > 0, so we have
√
1+x−
x
√
1−x
√
=
=
=
=
Since lim
√
x→0
1 + x = 1, and lim
√
x→0
1+x−
x
√
1−x
√
√
1+x+ 1−x
√
·√
1+x+ 1−x
x·
1 + x − (1 − x)
√
√
1+x+ 1−x
x·
√
√
2x
√
1+x+ 1−x
2
√
.
1+x+ 1−x
1 − x = 1, so by the additivity property of the limit, we know
that
√
√
lim [ 1 + x + 1 − x] = 2.
x→0
Then by the quotient rule of the limit, we know that
lim √
x→0
2
2
√
= = 1.
2
1+x+ 1−x
2. [Page 145, Problem 1] Let f be a polynomial of degree n, say f (x) =
n
X
ck xk , such that the first
k=0
and last coefficients c0 and cn have opposite signs, that is, c0 · cn < 0. Prove that f (x) = 0 for at least
one positive x.
1
2
MINGFENG ZHAO
Proof. Since c0 ·cn < 0, so we have two cases: Case I: If c0 > 0, and cn < 0, then we have f (0) = c0 > 0,
but since cn < 0, so
lim f (x) = −∞.
x→∞
So there exists some X > 0 such that whenever x ≥ X, we have
f (x) ≤ −1.
In particular, we get f (X) ≤ −1 < 0. Then by the Bolcano’s Theorem, we know that there exists
some x0 ∈ (0, X), we have
f (x0 ) = 0.
Case II: If c0 < 0, and cn > 0, then we have f (0) = c0 < 0, but since cn > 0, so
lim f (x) = ∞.
x→∞
So there exists some X > 0 such that whenever x ≥ X, we have
f (x) ≥ 1.
In particular, we get f (X) ≥ 1 > 0. Then by the Bolcano’s Theorem, we know that there exists
some x0 ∈ (0, X), we have
f (x0 ) = 0.
3. [Page 149, Problem 4] Show that f (x) = x3 is strictly monotonic in R, and compute the inverse
function g of f .
Proof. Claim I: f (x) = x3 is strictly monotonic in R.
For any x < y ∈ R and fix, then we know that y − x > 0, and
f (y) − f (x)
= y 3 − x3
SOLUTION OF HW7
=
3
(y − x)(y 2 + xy + x2 ).
If x ≥ 0, then y > x ≥ 0, which implies that y 2 > 0, xy ≥ 0 and x2 ≥ 0. So we have
f (y) − f (x) ≥ (y − x)y 2 > 0.
If x < 0, then x2 > 0, and
1
3
y 2 + xy + x2 = y 2 + xy + x2 + x2 =
4
4
1
y+ x
2
2
3
3
+ x2 ≥ x2 > 0.
4
4
So we have
3
f (y) − f (x) ≥ (y − x) · x2 > 0.
4
In a summary, we have f (y) − f (x) > 0. By the arbitrary of x < y ∈ R, we get f (y) > f (x) for all
x < y ∈ R, that is, f is strictly monotonic in R.
1
Claim II: g(x) = x 3 for all x ∈ R is the inverse function of f .
Since lim
y→±∞
x3 = ±∞, and f (x) = x3 is continuous on R, then the range of f (x) = x3 is R. Since
1
g(x) = x 3 for all x ∈ R, then for all x ∈ R, we have
f (g(x))
1
= f (x 3 )
=
1
(x 3 )3
= x
g(f (x))
= g(x3 )
=
1
(x3 ) 3
= x.
1
Therefore, we know that g(x) = x 3 for all x ∈ R is the inverse function of f .
4
MINGFENG ZHAO
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu