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 £© Section
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 £© Some Limit Calculations
In this section we will discuss the use of Taylor polynomials puting certain types of
limits. Although this material could have been treated directly after Section , we have
saved it until now so as not to break into the development of Taylor series. To illustrate the
ideas of this section, we begin with two examples, the first of which is already well-known
to us.
Example Consider the problem of evaluating
sin(x)
lim .
x→0 x
The reason this limit presents a problem is that, although the function in question is a
quotient of two continuous functions, both the numerator and the denominator approach
0 as x approaches 0. Now from our work on Taylor polynomials we know that
sin(x) = x + o(x),
so
sin(x) x + o(x) o(x)
= = 1 + .
x x x
But, by definition,
o(x)
lim = 0.
x→0 x
Thus
sin(x) x + o(x) o(x)
lim = lim = lim 1 + = 1.
x→0 x x→0 x x→0 x
Example The limit
1 − cos(x)
lim
x→0 x2
presents the same type of problem. Using the Taylor polynomial of order 2 for cos(x), we
know that
x2
cos(x) = 1 −+ o(x2).
2
1
2 Some Limit Calculations Section
Hence
x2
1 −1 −+ o(x2)
1 − cos(x) 2
lim = lim
x→0 x2 x→0 x2
x2
− o(x2)
= lim 2
x→0 x2
1 o(x2)
= lim −
x→0 2 x2
1
= .
2
The p