Analysis Problems #3
Solutions
1. Suppose that f is continuous with f (0) < 1. Show that there exists some δ > 0 such
that f (x) < 1 for all −δ < x < δ. Hint: use the ε-δ definition for some suitable ε.
• Since ε = 1 − f (0) is positive, some δ > 0 exists such that
|x − 0| < δ
=⇒
|f (x) − f (0)| < ε
=⇒
f (x) − f (0) < 1 − f (0).
In other words, one has f (x) < 1 for all −δ < x < δ, as needed.
2. Show that the polynomial f (x) = x5 − 3x + 1 has three roots in the interval (−2, 2).
Hint: you might wish to compute the values of f at the points ±2, ±1 and 0.
• Being a polynomial, f is continuous on the closed interval [−2, −1] and we also have
f (−2) = −25 < 0,
f (−1) = 3 > 0.
Thus, f must have a root in (−2, −1) by Bolzano’s theorem. Using the facts that
f (0) = 1 > 0,
f (1) = −1 < 0,
f (2) = 27 > 0,
we similarly find that a second root exists in (0, 1) and a third root exists in (1, 2).
3. Suppose that f is continuous with |f (x)| ≤ 1 for all x ∈ R. Show that there exists
some c ∈ R such that f (c) = c. Hint: apply Bolzano’s theorem on a suitable interval.
• Being the difference of two continuous functions, g(x) = f (x) − x is continuous with
g(2) = f (2) − 2 ≤ 1 − 2 < 0,
g(−2) = f (−2) + 2 ≥ −1 + 2 > 0.
According to Bolzano’s theorem then, some c ∈ (−2, 2) exists such that g(c) = 0.
4. Suppose that f is continuous on [0, 1] and that 0 < f (x) < 1 for all x ∈ [0, 1]. Show
that there exists some 0 < c < 1 such that f (c) = c.
• Let g(x) = f (x) − x for all x ∈ [0, 1]. Being the difference of two continuous functions,
g is then continuous on the closed interval [0, 1]. Once we now note that
g(0) = f (0) > 0,
g(1) = f (1) − 1 < 0,
we may use Bolzano’s theorem to conclude that g(c) = 0 for some c ∈ (0, 1). This also
implies that f (c) = c for some 0 < c < 1, as needed.
5. Compute each of the following limits:
L = lim
x→1
x3 − 4x2 − 4x − 1
,
2x3 + 3x2 + 3x − 2
M = lim
x→1
x3 − 4x2 + 4x − 1
.
2x3 − 3x2 + 3x − 2
• Since rational functions are continuous wherever defined, one easily gets
L = lim
x→1
x3 − 4x2 − 4x − 1
1−4−4−1
8
4
=
=− =−
3
2
2x + 3x + 3x − 2
2+3+3−2
6
3
by simple substitution. For the second limit, the substitution x = 1 gives 0/0, so a
factor of x − 1 should be present in both the numerator and the denominator. Using
division of polynomials to verify this, we now find that
M = lim
x→1
x3 − 4x2 + 4x − 1
(x − 1)(x2 − 3x + 1)
=
lim
.
2x3 − 3x2 + 3x − 2 x→1 (x − 1)(2x2 − x + 2)
Since rational functions are continuous wherever defined, this also implies that
M = lim
x→1
x2 − 3x + 1
1−3+1
1
=
=− .
2
2x − x + 2
2−1+2
3