This is a really cool trick I only recently learned. Evaluating
limits has useful applications, and this little trick allows the
evaluation of limits which, on the surface, seem impossible. Consider
x→0lim2x42cos2x−2+4x2.
The denominator poses a problem, as limx→02x4=0, therefore
the entire limit is of the form 0−2, which suffice it to
say is not acceptable. It isn’t one of the forms to which L’Hôpital’s
rule applies.
We expect the limit to be evaluable, however, by inspecting the graph.
However, there is another way to evaluate this limit, though it involves
making it appear very complex. By expanding cos2x using the
Maclaurin series for cosx, the limit can be expanded.
=====limx→02x42∑k=0∞(2k)!(−1)k(2x)2k−2+4x2limx→02x44x2−2+2[1−2!(2x)2+4!(2x)4−6!(2x)6+⋯]limx→02x44x2−2+2−2!2(2x)2+4!2(2x)4−6!2(2x)6+⋯limx→02x44!2(2x)4−6!2(2x)6+⋯limx→02x42x4⋅[4!24−6!26x2+8!28x4−⋯]2416=32
The remaining terms from the Maclaurin series are eliminated because
they tend to zero when the limit is evaluated, which leaves us with the
expected value of 32. That is unqualifiedly cool.
This is a solution for §9.4 Exercise #9 from Calculus: Early Transcendentals.