Taylor series and little-oh notation

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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Don't you need some bounds on $f'''(\xi)$? –  Srivatsan Sep 9 '11 at 20:27
Possible duplicate: Little o notation and series –  Henning Makholm Sep 9 '11 at 20:32
Just a guess, but I think this exercise is only intended as a language quiz. The student may be expected to write $o(x^2)$ without worrying about the actual behaviour of the series. –  André Nicolas Sep 9 '11 at 20:39
But when I try to get an expression for f ′′′ (ξ) I get something very long and ugly. Therefore I just called it a constant. f ′′′ (ξ) is a constant, so I have problem with seeing the mistake? –  Brugerfugl Sep 9 '11 at 20:51
There is no mistake, it is well done. Calculation of $f'''(x)$ is not that bad, and we don't even need to do the entire calculation to see that the third derivative is bounded in a neighbourhood of $0$. An explicit bound is not needed. –  André Nicolas Sep 9 '11 at 21:48