One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor?
$$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$
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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