One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor?
limx→0tan(x)−xx3
Answer
The statement tan(x)−xx3→c as x→0 is equivalent to
tan(x)=x+cx3+o(x3) as x→0, so this is a statement about a
Taylor polynomial of tan(x), and I'm not sure what would count as doing
that "without Taylor". However, one thing you could do is start from sin(x)=x+o(x)
integrate to get cos(x)=1−x2/2+o(x2)
then sec(x)=11−x2/2+o(x2)=1+x2/2+o(x2)
sec2(x)=(1+x2/2+o(x2))2=1+x2+o(x2)
and integrate again to get
tan(x)=x+x3/3+o(x3)
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