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.
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 $-{1\over30}$.
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 $$\tan (\sin (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+O\left(x^8\right)$$ $$\sin (\tan (x))-\tan (\sin (x))=-\frac{x^7}{30}+O\left(x^8\right)$$
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.
The function is of an indeterminate form, so use l'Hopital's Rule $7$ times.
Note that
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At $x=0$, this is equal to $-168$. Taking the derivative of the denominator $7$ times, we get $7!=5040$. Thus, the answer is $$ - \dfrac {168}{5040} = \boxed {- \dfrac {1}{30}}. $$
I already have a solution which uses some algebraic manipulation combined with Taylor series expansions, but I think it is worthwhile to find a neat solution based on L'Hospital's Rule.
My idea is to find a suitable approximation for both the expressions $\tan(\sin x)$ and $\sin(\tan x)$. I start with a simple approximation and improve on it further. After some guess work and manipulation with various limits I found that the expression $(2x - \sin x)$ acts as a very good approximation for both the expressions. This is justified by the existence of following limits \begin{align} \lim_{x \to 0}\frac{\tan(\sin x) - (2x - \sin x)}{x^{5}} = A\tag{1}\\ \lim_{x \to 0}\frac{\sin(\tan x) - (2x - \sin x)}{x^{5}} = B\tag{2} \end{align} The first limit is calculated as follows \begin{align} A &= \lim_{x \to 0}\frac{\tan(\sin x) + \sin x - 2x}{x^{5}}\notag\\ &= \lim_{x \to 0}\frac{\tan(\sin x) + \sin x - 2x}{\sin^{5}x}\cdot\frac{\sin^{5}x}{x^{5}}\notag\\ &= \lim_{x \to 0}\frac{\tan(\sin x) + \sin x - 2x}{\sin^{5}x}\notag\\ &= \lim_{t \to 0}\frac{\tan t + t - 2\arcsin t}{t^{5}}\notag\\ &= \lim_{t \to 0}\dfrac{\sec^{2}t + 1 - \dfrac{2}{\sqrt{1 - t^{2}}}}{5t^{4}}\text{ (by LHR)}\notag\\ &= \frac{1}{5}\lim_{t \to 0}\frac{(1 + \sec^{2}t)\sqrt{1 - t^{2}} - 2}{t^{4}\sqrt{1 - t^{2}}}\notag\\ &= \frac{1}{5}\lim_{t \to 0}\frac{(1 + \sec^{2}t)\sqrt{1 - t^{2}} - 2}{t^{4}}\notag\\ &= \frac{1}{5}\lim_{t \to 0}\frac{(2 + \tan^{2}t)^{2}(1 - t^{2}) - 4}{t^{4}\{(1 + \sec^{2}t)\sqrt{1 - t^{2}} + 2\}}\notag\\ &= \frac{1}{20}\lim_{t \to 0}\frac{(2 + \tan^{2}t)^{2}(1 - t^{2}) - 4}{t^{4}}\notag\\ &= \frac{1}{20}\lim_{t \to 0}\frac{4\tan^{2}t + \tan^{4}t - 4t^{2} - 4t^{2}\tan^{2}t - t^{2}\tan^{4}t}{t^{4}}\notag\\ &= \frac{1}{20}\left(\lim_{t \to 0}\frac{4(\tan^{2}t - t^{2})}{t^{4}} + \frac{\tan^{4}t}{t^{4}} - 4\frac{\tan^{2}t}{t^{2}} - t^{2}\frac{\tan^{4}t}{t^{4}}\right)\notag\\ &= \frac{1}{5}\lim_{t \to 0}\frac{\tan^{2} - t^{2}}{t^{4}} - \frac{3}{20}\notag\\ &= -\frac{3}{20} + \frac{1}{5}\lim_{t \to 0}\frac{\tan t - t}{t^{3}}\cdot\lim_{t \to 0}\frac{\tan t + t}{t}\notag\\ &= -\frac{3}{20} + \frac{2}{5}\lim_{t \to 0}\frac{\tan t - t}{t^{3}}\notag\\ &= -\frac{3}{20} + \frac{2}{5}\lim_{t \to 0}\frac{\sec^{2}t - 1}{3t^{2}}\text{ (by LHR)}\notag\\ &= -\frac{3}{20} + \frac{2}{15} = -\frac{1}{60}\notag \end{align} For the second limit we proceed in similar manner \begin{align} B &= \lim_{x \to 0}\frac{\sin(\tan x) + \sin x - 2x}{x^{5}}\notag\\ &= \lim_{x \to 0}\frac{\sin(\tan x) + \sin x - 2x}{\tan^{5}x}\cdot\frac{\tan^{5}x}{x^{5}}\notag\\ &= \lim_{x \to 0}\frac{\sin(\tan x) + \sin x - 2x}{\tan^{5}x}\notag\\ &= \lim_{t \to 0}\dfrac{\sin t + \dfrac{t}{\sqrt{1 + t^{2}}} - 2\tan^{-1}t}{t^{5}}\notag\\ &= \lim_{t \to 0}\dfrac{\cos t + \dfrac{1}{(1 + t^{2})^{3/2}} - \dfrac{2}{1 + t^{2}}}{5t^{4}}\text{ (via LHR)}\notag\\ &= \frac{1}{5}\lim_{t \to 0}\dfrac{\cos t(1 + t^{2})^{3/2} + 1 - 2\sqrt{1 + t^{2}}}{t^{4}(1 + t^{2})^{3/2}}\notag\\ &= \frac{1}{5}\lim_{t \to 0}\dfrac{\{(1 + t^{2})\cos t - 2\}\sqrt{1 + t^{2}} + 1}{t^{4}}\notag\\ &= \frac{1}{5}\lim_{t \to 0}\dfrac{\{(1 + t^{2})\cos t - 2\}^{2}(1 + t^{2}) - 1}{t^{4}[\{(1 + t^{2})\cos t - 2\}\sqrt{1 + t^{2}} - 1]}\notag\\ &= -\frac{1}{10}\lim_{t \to 0}\dfrac{\{(1 + t^{2})^{2}\cos^{2}t + 4 - 4(1 + t^{2})\cos t\}(1 + t^{2}) - 1}{t^{4}}\notag\\ &= -\frac{1}{10}\lim_{t \to 0}\dfrac{\{(1 + t^{4} + 2t^{2})(1 - \sin^{2}t) + 4 - 4(1 + t^{2})(1 - 2\sin^{2}(t/2))\}(1 + t^{2}) - 1}{t^{4}}\notag\\ &= -\frac{1}{10}\lim_{t \to 0}\left(\frac{8\sin^{2}(t/2) - \sin^{2}t - t^{2}}{t^{4}}\right.\notag\\ &\,\,\,\,\,\,\,+ \dfrac{16\sin^{2}(t/2) - 3\sin^{2}t}{t^{2}}\notag\\ &\,\,\,\,\,\,\,+ (-1 - 3\sin^{2}t + 8\sin^{2}(t/2))\notag\\ &\,\,\,\,\,\,\,+ \left.\dfrac{(1 - \sin^{2}t)t^{6}}{t^{4}}\right)\notag\\ &= -\frac{1}{10}\lim_{t \to 0}\left(\frac{8\sin^{2}(t/2) - \sin^{2}t - t^{2}}{t^{4}} + 4 - 3 - 1 + 0\right)\notag\\ &= -\frac{1}{10}\lim_{t \to 0}\left(\frac{8\sin^{2}(t/2) - 2\sin^{2}t + \sin^{2}t - t^{2}}{t^{4}}\right)\notag\\ &= -\frac{1}{10}\lim_{t \to 0}\left(\frac{8\sin^{2}(t/2) - 8\sin^{2}(t/2)\cos^{2}(t/2)}{t^{4}} + \frac{\sin t - t}{t^{3}}\cdot\frac{\sin t + t}{t}\right)\notag\\ &= -\frac{1}{10}\lim_{t \to 0}\left(\frac{8\sin^{4}(t/2)}{t^{4}} - \frac{1}{3}\right)\notag\\ &= -\frac{1}{10}\left(\frac{1}{2} - \frac{1}{3}\right) = -\frac{1}{60}\notag \end{align} From these limit $(1), (2)$ we easily get \begin{align} \lim_{x \to 0}\frac{\tan(\sin x) + \sin x - 2x}{\sin^{5}x} &= \lim_{x \to 0}\frac{\tan(\sin x) + \sin x - 2x}{x^{5}}\cdot\frac{x^{5}}{\sin^{5}x} = -\frac{1}{60}\tag{3}\\ \lim_{x \to 0}\frac{\sin(\tan x) + \sin x - 2x}{\tan^{5}x} &= \lim_{x \to 0}\frac{\sin(\tan x) + \sin x - 2x}{x^{5}}\cdot\frac{x^{5}}{\tan^{5}x} = -\frac{1}{60}\tag{4} \end{align} From these equations we get the next level of approximations \begin{align} \tan(\sin x) &\approx 2x - \sin x - \frac{\sin^{5}x}{60}\notag\\ \sin(\tan x) &\approx 2x - \sin x - \frac{\tan^{5}x}{60}\notag \end{align} Our claim is that the following limits exist \begin{align} \lim_{x \to 0}\dfrac{\sin(\tan x) - 2x + \sin x + \dfrac{\tan^{5}x}{60}}{x^{7}} &= C\tag{5}\\ \lim_{x \to 0}\dfrac{\tan(\sin x) - 2x + \sin x + \dfrac{\sin^{5}x}{60}}{x^{7}} &= D\tag{6} \end{align} Let the desired limit in question be $L$ i.e. $$L = \lim_{x \to 0}\frac{\sin(\tan x) - \tan(\sin x)}{x^{7}}\tag{7}$$ Subtracting $(6)$ from $(5)$ we get $$L + \frac{1}{60}\lim_{x \to 0}\frac{\tan^{5}x - \sin^{5}x}{x^{7}} = C - D\tag{8}$$ Now it is easy to see that \begin{align} E &= \lim_{x \to 0}\frac{\tan^{5}x - \sin^{5}x}{x^{7}}\notag\\ &= \lim_{x \to 0}\frac{\tan x - \sin x}{x^{3}}\cdot\frac{\tan^{4}x + \tan^{3}x\sin x + \tan^{2}x\sin^{2}x + \tan x\sin^{3}x + \sin^{4}x}{x^{4}}\notag\\ &= 5\lim_{x \to 0}\frac{\sin x(1 - \cos x)}{x^{3}\cos x}\notag\\ &= 5\lim_{x \to 0}\frac{\sin x}{x}\cdot\frac{1 - \cos x}{x^{2}}\notag\\ &= \frac{5}{2}\notag \end{align} From equation $(8)$ we can see that $L = C - D - (1/24)$. It remains to evaluate $C$ and $D$ which can be done in a manner similar to the evaluation of $A$ and $B$. We will provide here the calculation for the simpler limit $D$ and let readers evaluate $C$ by themselves. We can thus proceed as follows \begin{align} D &= \lim_{x \to 0}\dfrac{\tan(\sin x) - 2x + \sin x + \dfrac{\sin^{5}x}{60}}{x^{7}}\notag\\ &= \lim_{x \to 0}\dfrac{\tan(\sin x) - 2x + \sin x + \dfrac{\sin^{5}x}{60}}{\sin^{7}x}\cdot\frac{\sin^{7}x}{x^{7}}\notag\\ &= \lim_{x \to 0}\dfrac{\tan(\sin x) - 2x + \sin x + \dfrac{\sin^{5}x}{60}}{\sin^{7}x}\notag\\ &= \lim_{t \to 0}\dfrac{\tan t - 2\sin^{-1}t + t + \dfrac{t^{5}}{60}}{t^{7}}\notag\\ &= \frac{1}{60}\lim_{t \to 0}\dfrac{60\tan t - 120\sin^{-1}t + 60t + t^{5}}{t^{7}}\notag\\ &= \frac{1}{420}\lim_{t \to 0}\dfrac{60\sec^{2}t - \dfrac{120}{\sqrt{1 - t^{2}}} + 60 + 5t^{4}}{t^{6}}\text{ (via LHR)}\notag\\ &= \frac{1}{420}\lim_{t \to 0}\dfrac{\{60(2 + \tan^{2}t) + 5t^{4}\}\sqrt{1 - t^{2}} - 120}{t^{6}\sqrt{1 - t^{2}}}\notag\\ &= \frac{1}{420}\lim_{t \to 0}\dfrac{\{60(2 + \tan^{2}t) + 5t^{4}\}\sqrt{1 - t^{2}} - 120}{t^{6}}\notag\\ &= \frac{1}{420}\lim_{t \to 0}\dfrac{\{60(2 + \tan^{2}t) + 5t^{4}\}^{2}(1 - t^{2}) - 120^{2}}{t^{6}[\{60(2 + \tan^{2}t) + 5t^{4}\}\sqrt{1 - t^{2}} + 120]}\notag\\ &= \frac{1}{420}\cdot\frac{1}{240}\lim_{t \to 0}\dfrac{\{60(2 + \tan^{2}t) + 5t^{4}\}^{2}(1 - t^{2}) - 120^{2}}{t^{6}}\notag\\ &= \frac{1}{420}\cdot\frac{1}{240}\lim_{t \to 0}\left(\frac{60^{2}\tan^{4}t + 1200t^{4} + 120^{2}(\tan^{2}t - t^{2} - t^{2}\tan^{2}t)}{t^{6}}\right.\notag\\ &\,\,\,\,\,\,\,\,+\frac{600t^{4}\tan^{2}t - 60^{2}t^{2}\tan^{4}t - 1200t^{6}}{t^{6}}\notag\\ &\,\,\,\,\,\,\,\,+\left. \frac{- 600t^{6}\tan^{2}t + 25t^{8} - 25t^{10}}{t^{6}}\right)\notag\\ &= \frac{1}{420}\cdot\frac{1}{240}\lim_{t \to 0}\left(1200\cdot\frac{3\tan^{4}t + t^{4} + 12(\tan^{2}t - t^{2} - t^{2}\tan^{2}t)}{t^{6}} - 4200 + 0\right)\notag\\ &= -\frac{1}{24} + \frac{1}{84}\lim_{t \to 0}\frac{3\tan^{4}t + t^{4} + 12(\tan^{2}t - t^{2} - t^{2}\tan^{2}t)}{t^{6}}\notag\\ &= -\frac{1}{24} + \frac{1}{84}\lim_{t \to 0}\frac{3(\tan^{4} - t^{4}) - 12t^{2}(\tan^{2}t - t^{2}) + 12(\tan t - t)^{2} + 8t(3\tan t- 3t- t^{3})}{t^{6}}\notag\\ &= -\frac{1}{24} + \frac{1}{84}\lim_{t \to 0}\left(3\cdot\frac{\tan t - t}{t^{3}}\cdot\frac{\tan^{3}t + t\tan^{2}t + t^{2}\tan t + t^{3}}{t^{3}}\right.\notag\\ &\,\,\,\,\,\,\,\, -12\cdot \frac{\tan t - t}{t^{3}}\cdot\frac{\tan t + t}{t}\notag\\ &\,\,\,\,\,\,\,\, + 12\cdot\left(\frac{\tan t - t}{t^{3}}\right)^{2}\notag\\ &\,\,\,\,\,\,\,\, + \left. 8\cdot\frac{3\tan t - 3t - t^{3}}{t^{5}}\right)\notag\\ &= -\frac{1}{24} - \frac{2}{63} + \frac{2}{21}\lim_{t \to 0}\frac{3\tan t - 3t - t^{3}}{t^{5}}\notag\\ &= -\frac{37}{504} + \frac{2}{105}\lim_{t \to 0}\frac{3\sec^{2}t - 3 - 3t^{2}}{t^{4}}\text{ (via LHR)}\notag\\ &= -\frac{37}{504} + \frac{2}{35}\lim_{t \to 0}\frac{\tan t - t}{t^{3}}\cdot\frac{\tan t + t}{t}\notag\\ &= -\frac{37}{504} + \frac{4}{105} = -\frac{89}{2520}\notag \end{align} A similar but somewhat lengthy calculation shows that the limit $C = -17/630$ and hence desired limit $L = -1/30$.
In the above solution we make use of the following limits $$\lim_{x \to 0}\frac{\sin x - x}{x^{3}} = -\frac{1}{6},\,\lim_{x \to 0}\frac{\tan x - x}{x^{3}} = \frac{1}{3}$$ which are easily evaluated via one application of L'Hospital's rule.
Needless to say that the above approach uses heavy algebraic manipulation and expresses the given problem as equivalent of 4 tough limit problems $(1), (2), (5), (6)$. Moreover for calculation of the desired limit $L$ we only need to calculate limits $C, D$. The limits $A, B$ are needed to guess the limit expression used for $C, D$. It is thus evident that after a certain point Taylor's series approach is far more favorable compared to the L'Hospital's rule. Moreover L'Hospital's rule is effective only when it is combined with a reasonable level of algebraic manipulation. One should always avoid repeated application of LHR in successive steps.
Also note that we did not use the approximation $$2x - \sin x - \frac{x^{5}}{60}$$ for both $\sin(\tan x)$ and $\tan(\sin x)$ based on limits $(1)$ and $(2)$. Doing this would have made the difference $\sin(\tan x) - \tan(\sin x)$ as $0$. Instead the term $x^{5}$ was replaced very smartly with $\sin^{5}x$ in one case and $\tan^{5}x$ in another case because of the limit $E$ which is related to the difference $\tan^{5}x - \sin^{5}x$.
A final point which I wish to make is that this approach starts with the initial approximation $(2x - \sin x)$ which is fairly accurate. It may be possible that there is a simpler approximation which makes the calculations much easier.
Why don't you try L'Hopital's rule? You have an indeterminate form $\frac{0}{0} $
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.
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 $7!=5040$.
