Analysis Problems #2
Solutions
1. Let f be a function which is defined on the interval [0, 2]. Show that
inf f (x) ≥ inf f (x).
0≤x≤1
0≤x≤2
• We have to show that inf A ≥ inf B, where
A = {f (x) : 0 ≤ x ≤ 1} and B = {f (x) : 0 ≤ x ≤ 2}.
Since A ⊂ B, this is a simple consequence of Theorem 1.27.
2. Compute each of the following limits:
2x3 − 5x − 6
,
x→2
x−2
lim
x3 − 3x + 2
.
x→1
(x − 1)2
lim
• When it comes to the first limit, division of polynomials gives
2x3 − 5x − 6
= lim (2x2 + 4x + 3) = 2 · 22 + 4 · 2 + 3 = 19
x→2
x→2
x−2
lim
because x ̸= 2 here and since polynomial functions are known to be continuous.
• When it comes to the second limit, a similar computation gives
x3 − 3x + 2
x3 − 3x + 2
lim
= lim 2
= lim (x + 2) = 3.
x→1
x→1 x − 2x + 1
x→1
(x − 1)2
3. Let f be the function defined by
{
f (x) =
7 − 2x
3x − 3
if x ∈ Q
if x ∈
/Q
}
.
Use the ε-δ definition of limits to show that lim f (x) = 3.
x→2
• Let ε > 0 be given and note that
{
2|2 − x|
|f (x) − 3| =
3|x − 2|
if x ∈ Q
if x ∈
/Q
}
≤ 3|x − 2|.
Taking δ = ε/3, we conclude that
|x − 2| < δ
=⇒
|f (x) − 3| ≤ 3|x − 2| < 3δ = ε.
4. Evaluate the limit lim f (x) when f is the function defined by
x→1
{
f (x) =
1 + 3x
5
if x ̸= 1
if x = 1
}
.
• Since x ̸= 1 whenever x → 1, one easily finds that
lim f (x) = lim (1 + 3x) = 1 + 3 = 4
x→1
x→1
because limits of polynomials can be computed by simple substitution.
5. Use the ε-δ definition of limits to show that lim x2 = 4.
x→2
• Let ε > 0 be given and note that the triangle inequality gives
|x − 2| < δ
=⇒
|x + 2| = |x − 2 + 4| ≤ |x − 2| + 4 < δ + 4.
Taking δ = min{1, ε/6}, we conclude that
|x − 2| < δ
=⇒
|x2 − 4| = |x + 2| · |x − 2| ≤ (δ + 4) · δ ≤ 5 ·
ε
< ε.
6