James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

192 Chapter 3 Differentiation Rules
length of an arc without resorting to geometric intuition as we did here.) Therefore we
have
, sin
cos
so
cos , sin
, 1
We know that lim l 0 1 − 1 and lim l 0 cos − 1, so by the Squeeze Theorem, we have
sin
lim − 1
l0 1
We multiply numerator and denominator
by cos 1 1 in order to put the
function in a form in which we can use
the limits we know.
But the function ssin dy is an even function, so its right and left limits must be equal.
Hence, we have
sin
lim − 1
l 0
so we have proved Equation 2.
We can deduce the value of the remaining limit in (1) as follows:
cos 2 1
lim − lim
l 0
l 0S cos 2 1 ? cos 1 1 cos
− lim
cos 1 1D 2 2 1
l 0 scos 1 1d
− lim
l 0
2sin 2
scos 1 1d − 2lim
l 0S
sin
− 2lim ? lim
l 0 l 0
sin
cos 1 1
sin ?
− 21 ?S 0
1 1 1D − 0 (by Equation 2)
sin
cos 1 1D
cos 2 1
3 lim − 0
l 0
If we now put the limits (2) and (3) in (1), we get
cos h 2 1
sin h
f 9sxd − lim sin x ? lim 1 lim cos x ? lim
h l 0 h l 0 h
h l 0 h l 0 h
− ssin xd ? 0 1 scos xd ? 1 − cos x
So we have proved the formula for the derivative of the sine function:
4
d
ssin xd − cos x
dx
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