Section 1.3: The notion of limit
(continued)
For the function
f
(
x
) =
x
sin 1/
x
(defined for all
x
0) do we
≠
have lim
x
→
0
f
(
x
) = 0?
0.4
0.2
0.2
0.4
0.2
0.1
0.1
0.2
0.3
0.4
..?.
.
Yes.
What’s Eve’s strategy?
..?.
.
Eve wins by taking
δ
=
ε
.
Check: For every
x
satisfying 0 < 
x
– 0 <
, we have

x
sin 1/
x
– 0 = 
x
sin 1/
x
 = 
x
 sin 1/
x

≤

x
 1 = 
x
 = 
x
– 0
<
=
.
(Note that this proof makes use of the handy fact that 
ab
 =
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a
 
b
, which you’ll prove in the next homework
assignment.
Another handy fact about absolute values is
that 
a
+
b

≤

a
 + 
b
; this is called the
triangle inequality
.)
Stewart’s paraphrase of the definition of limits: We say
lim
x
→
a
f
(
x
) =
L
if the values of
f
(
x
) tend to get closer and
closer to the number
L
as
x
gets closer and closer to the
number
a
(from either side of
a
) but
x
≠
a
.
But what does “closer and closer” mean?
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 Fall '11
 Staff
 Calculus, Inequalities, Norm, true assertion

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