One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor?
lim
Answer
The statement \dfrac{\tan(x)-x}{x^3} \to c as x \to 0 is equivalent to
\tan(x) = x + c x^3 + o(x^3) as x \to 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 - x^2/2 + o(x^2) then \sec(x) = \frac{1}{1-x^2/2 + o(x^2)} = 1 + x^2/2 + o(x^2) \sec^2(x) = \left(1 + x^2/2 + o(x^2)\right)^2 = 1 + x^2 + o(x^2) and integrate again to get
\tan(x) = x + x^3/3 + o(x^3)
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