a. Consider the series \begin{equation} e^{\tan(x)}=1+x+\frac{x^{2}}{2!}+\frac{3x^{3}}{3!}+\frac{9x^{4}}{4!}+\dots\qquad (|x|\leq\pi/2) \end{equation} Retaining three terms in the series, estimate the remaining series using o-notation with the best integer value possible, as $x\to 0$.
b. Repeat the same problem using O-notation and the series \begin{equation} \ln(\tan(x))=\ln(x)+\frac{x^{2}}{3}+\frac{7x^{4}}{90}+\frac{62x^{6}}{2835}+\dots\qquad (0<|x|<\pi/2). \end{equation}
In general, we write \begin{equation} f(x)=O(g(x))\qquad (x\to x^{*}) \end{equation} when there is a constant C and a neighborhood of $x^{*}$ such that $\left | f(x) \right |\leq C\left | g(x) \right |$ in that neighborhood. Similarly, \begin{equation} f(x)=o(g(x))\qquad (x\to x^{*}) \end{equation} means that $\lim_{x\to > x^{*}}\frac{f(x)}{g(x)}=0$.
(Source: Numerical Analysis: Mathematics of Scientific Computing)
My answer: We start with a, we get \begin{equation} e^{\tan(x)}-\left (1+x+\frac{x^{2}}{2!} \right )=\frac{3x^{3}}{3!}+\frac{9x^{4}}{4!}+\dots=o(x^{n}) \end{equation} which is \begin{equation} \lim_{x\to 0}\frac{\frac{3x^{3}}{3!}+\frac{9x^{4}}{4!}+\dots}{x^{n}}=\lim_{x\to 0}\frac{3x^{3-n}}{3!}+\frac{9x^{4-n}}{4!}+\dots=0 \end{equation} Now I'm stuck. I've no idea what I'm going to do the next step. I'd also be appreciated if you explain me about the other problem (b), please.
