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Consider the series

$e^{\tan(x)} = 1 + x + \dfrac{x^{2}}{2!} + \dfrac{3x^{3}}{3!} + \dfrac{9x^{4}}{4!} + \ldots $

Retaining three terms in the series, estimate the remaining series using "Little-$o$" notation with the best integer value possible, as $x\to 0$.

I have tried to solve this problem.

$f(x) = e^{\tan(x)} = 1 + x + \dfrac{x^{2}}{2!} + E_{2}(x)$

Because E_{2}(x) = \dfrac{1}{6} f^{'''}(\xi)x^{3} I can conclude that:

$Cx^{3}\leq E_{2}(x)$ where $C$ is a constant

I now find the solution to the problem to be:

$E_{2}(x)=o(x^{\alpha})=o(x^{2})$

$\alpha = 2$ is true, because following should be true:

$\lim_{x\to 0}\dfrac{x^{3}}{x^{\alpha}}=0$ (Here I have used the little-oh definition)

Is it correct?

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