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**Unformatted text preview: **mits as x → 0.
Example.
Find lim x→0 sin(x)−x
.
x2 First notice that this is an indeterminant limit of the type 0 .
0
We know that if f (x) = sin(x), then T1,0 (x) = T 2, 0(x) = x. We will assume that we are
working with T2,0 (x). Then Taylor’s Theorem shows that for any x ∈ [−1, 1], there exists
a c between 0 and x such that
| sin(x) − x |=| − cos(c) 3
1
x |≤ | x |3
3!
6 since | − cos(c) |≤ 1 no matter where c is located. This inequality is equivalent to
−1
1
| x |3 ≤ sin(x) − x ≤ | x |3 .
6
6
If x = 0, we can divide all of the terms by x2 to get that for x ∈ [−1, 1] and x = 0
− | x |3
sin(x) − x
| x |3
≤
≤
6x2
x2
6x2
or equivalently that
−|x|
sin(x) − x
|x|
≤
≤
.
6
x2
6
We also know that −|x|
|x|
= lim
=0
x→0
x→0 6
6
The Squeeze Theorem shows that
lim sin(x) − x
.
x→0
x2
lim The technique we outlined in the previous example can be used in much more generality.
We will need the following observation:
Suppose that f (k+1) (x) is a continuous function on [...

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