How find this limit $\lim\limits_{x\to 0^{+}}\frac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$ 
Find the limit
$$\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$$

My attempt: Since
$$\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+o(x^7)$$
$$\tan{x}=x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+\dfrac{1}{63}x^7+o(x^3)$$
So
$$\sin{(\tan{x})}=\tan{x}-\dfrac{1}{3!}(\tan{x})^3+\dfrac{1}{5!}(\tan{x})^5-\dfrac{1}{7!}(\tan{x})^7+o(x^7)$$
Though this method might solve, I think this problem has nicer methods. Thanks.
 A: For sure, L'Hopital's rule would be useful. But you could also use each Taylor expansion of $\sin(x)$ and $\tan(x)$ to expand $\sin (\tan (x))$ and $\tan (\sin (x))$. This would probably be tedious but it is doable (I made it). 
Using what you wrote, you should arrive to $$\sin (\tan (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55 x^7}{1008}+O\left(x^8\right)$$ and $$\tan (\sin (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+O\left(x^8\right)$$ $$\sin (\tan (x))-\tan (\sin (x))=-\frac{x^7}{30}+O\left(x^8\right)$$
A: If you want to do it with Taylor expansions it's probably a good idea to start by substituting $x = \arcsin t$, giving 
$$ \lim_{t \to 0} \frac{\sin\left(\frac t {\sqrt{1-t^2}}\right) - \tan t}{(\arcsin t)^7}$$
Now use the expansion $$ \frac t {\sqrt{1-t^2}} = \sum_{n\geq 0} \binom{ -1/2} n (-1)^n t^{2n+1} = \sum_{n\geq 0} \frac{(2n-1)!!}{n! 2^n} t^{2n+1}$$
and expand denominator and numerator to order $t^7$. This will still take some calculation but at least less than the naive method.
A: The function is of an indeterminate form, so use l'Hopital's Rule $7$ times. 
Note that 
. 
At $x=0$, this is equal to $-168$. Taking the derivative of the denominator $7$ times, we get $7!=5040$. Thus, the answer is $$ - \dfrac {168}{5040} = \boxed {- \dfrac {1}{30}}. $$
A: The interesting thing (which I cannot explain) is that if you have two odd functions
$$f(x)=x+a_3x^3+a_5 x^5+a_7 x^7+?x^9,\quad g(x)=x+b_3x^3+b_5 x^5+b_7 x^7+?x^9$$
with $f'(0)=g'(0)=1$ (or $=-1$) then they "commute up to order 5", i.e.,
$$f\bigl(g(x)\bigr)-g\bigl(f(x)\bigr)= ?x^7\qquad(x\to0)\ .$$
To prove this we do the computation for $f\bigl(g(x)\bigr)$:
$$\eqalign{f\bigl(g(x)\bigr)&=x+(a_3+b_3)x^3+(a_5+3a_3b_3+b_5)x^5 + \cr &\qquad\qquad(a_7+5a_5b_3+3a_3b_3^2+3a_3b_5+b_7)x^7\ +\ ?x^9\ .\cr}\tag{1}$$
We see that the coefficients of  $x^3$ and $x^5$ both are symmetric in $a$ and $b$, and in addition  $a_7+b_7$ will cancel when forming $f\bigl(g(x)\bigr)-g\bigl(f(x)\bigr)$. From inspection of $(1)$ we therefore can deduce that
$$f\bigl(g(x)\bigr)-g\bigl(f(x)\bigr)=\bigl(2(a_5 b_3-a_3 b_5)+3(a_3b_3^2-a_3^2 b_3)\bigr)x^7\ +\ ?x^9\ .$$
Inserting here the known coefficients for $\sin$ and $\tan$ we find that the limit in question is $-{1\over30}$.
A: Why don't you try L'Hopital's rule? You have an indeterminate  form $\frac{0}{0} $
A: I'm going to throw my hat in with L'Hopital's Rule.
The first non-zero derivative of the numerator at $x=0$ is the seventh;therefore, one can use L'Hopital's Rule until the denominator is a constant.
Wolfram Alpha Link
Therefore your limit is
$$
\lim_{x\rightarrow 0^+} \frac{\sin(\tan(x))-\tan(\sin(x))}{x^7}=\frac{-168}{5040}=-\frac{1}{30} 
$$
The value (viz. 0) of the lower-order derivatives can be easily checked in the above link using the same method as this one was computed (just with fewer derivative operators). Likewise, it is simple to check that $7!=5040$.
