Limits necessary to determine the derivatives of sinx and
cosx
On this page, we set the stage to derive the values of (sinx)′ and (cosx)′. This is particularly interesting to
those of you who are curious about how we determine such
derivatives for which we do not (yet) have a derivative rule.
You will need to know the values of the
following limits later in the semester, not only for this
application.
Unless you are told otherwise, you will not be expected to be
able to reproduce the following proofs, but it is certainly true
that working through the rest of this page, carefully, will
increase your mathematical understanding in many ways.
The two important limits (derived below):
limx→0sin(x)x=1andlimx→01−cos(x)x=0.
Notice that both of these
limits have the 00 indeterminate form, so we need to do
more work, but what work?
In the following video, we use geometry and the squeeze theorem to
derive the first of the two limits:
limx→0sin(x)x=1
Once we know the first limit, the second limit
limx→0cos(x)−1x=0
is fairly easy:
limx→0cos(x)−1x=limx→0(cos(x)−1x⋅cos(x)+1cos(x)+1)=limx→0(cos2(x)−1x(1+cos(x)))=limx→0(−sin2(x)x(1+cos(x)))=−limx→0(sin(x)x⋅sin(x)(1+cos(x)))=−(limx→0sin(x)x)⋅(limx→0sin(x)1+cos(x)).
We just showed that the first of the limits on the last line is 1,
and
the second limit is
sin(0)1+cos(0)=02=0,
so the product is 0.