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

Solution 1:

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 $-{1\over30}$.

Solution 2:

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 $$\tan (\sin (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+O\left(x^8\right)$$ $$\sin (\tan (x))-\tan (\sin (x))=-\frac{x^7}{30}+O\left(x^8\right)$$

Solution 3:

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.

Solution 4:

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 $$ - \dfrac {168}{5040} = \boxed {- \dfrac {1}{30}}. $$

Solution 5:

Why don't you try L'Hopital's rule? You have an indeterminate form $\frac{0}{0} $