"Calculate the limit using Taylor's formula". I know that the limit should be 1/30. Problem is how to prove it.
$\displaystyle
\lim_{x\rightarrow0}\frac{tan(sinx) - sin(tanx)}{x^7}
$
Let $\displaystyle f(x)=\tan(\sin(x))$ and $\displaystyle g(x)=\sin(\tan(x))$. Then
$\displaystyle f(0)=0$ ----$\displaystyle g(0)=0$
$\displaystyle f'(0)=1$ ----$\displaystyle g'(0)=1$
$\displaystyle f''(0)=0$ ----$\displaystyle g''(0)=0$
$\displaystyle f^{(3)}(0)=1$ ----$\displaystyle g^{(3)}(0)=1$
$\displaystyle f^{(4)}(0)=0$---- $\displaystyle g^{(4)}(0)=0$
$\displaystyle f^{(5)}(0)=-3$ ----$\displaystyle g^{(5)}(0)=-3$
$\displaystyle f^{(6)}(0)=0$ ----$\displaystyle g^{(6)}(0)=0$
$\displaystyle f^{(7)}(0)=-107$---- $\displaystyle g^{(7)}(0)=-275$
$\displaystyle \therefore\quad\tan(\sin(x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107x^7}{5040}\pm\cdots$
and $\displaystyle \sin(\tan(x))=x+\frac{x^3}{6}-\frac{x^5}{4}-\frac{275x^7}{5040}\pm\cdots$
$\displaystyle \begin{aligned}\therefore\quad&\lim_{x\to{0}}\frac {\tan(\sin(x))-\sin(\tan(x))}{x^7}\\
&=\lim_{x\to{0}}\frac{x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107x^7}{5040}\pm\cdots-\left(x+\frac{x^3}{6}-\frac{x^5}{4}-\frac{275x^7}{5040}\pm\cdots\right)}{x^7}\\
&=\lim_{x\to{0}}\frac{\frac{x^7}{30}\pm\cdots}{x^7 }\\
&=\frac{1}{30}
\end{aligned}$