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?