Evaluating $\lim_{x\to0}{{\left(\frac{\tan{(1+\tan x)}}{\tan{(1+\sin x)}}\right)}^{1/x^3}}$ I am working on $$\lim_{x\to0}{{\left(\frac{\tan{(1+\tan x)}}{\tan{(1+\sin x)}}\right)}^{1/x^3}}$$
I can see a successful strategy is to take $\ln$ first and then use L'Hospital rule for 3 times. However, I wonder if there are simpler methods.
Any ideas would be appreciated!
 A: By standard limits we have that
$${\left(\frac{\tan{(1+\tan x)}}{\tan{(1+\sin x)}}\right)}^{1/x^3}=\left[{\left(1+\left(\frac{\tan{(1+\tan x)}}{\tan{(1+\sin x)}}-1\right)\right)}^{\frac{1}{\frac{\tan{(1+\tan x)}}{\tan{(1+\sin x)}}-1}}\right]^{\frac{\frac{\tan{(1+\tan x)}}{\tan{(1+\sin x)}}-1}{x^3}} \to e^{\csc 2}$$
indeed
$${\left(1+\left(\frac{\tan{(1+\tan x)}}{\tan{(1+\sin x)}}-1\right)\right)}^{\frac{1}{\frac{\tan{(1+\tan x)}}{\tan{(1+\sin x)}}-1}} \to e$$
and
$$\frac{\frac{\tan{(1+\tan x)}}{\tan{(1+\sin x)}}-1}{x^3}=\frac1{\tan{(1+\sin x)}}\frac{\tan{(1+\tan x)-\tan{(1+\sin x)}}}{x^3}\to \frac12\frac{1+\tan^2 1}{\tan 1}=\csc 2$$
since
$$\frac{\tan{(1+\tan x)-\tan{(1+\sin x)}}}{x^3}=\frac{\frac{\tan 1+\tan(\tan x)}{1-\tan 1\tan (\tan x)}-\frac{\tan 1+\tan(\sin x)}{1-\tan 1\tan(\sin x)}}{x^3}=$$
$$=\frac{1+\tan^21}{(1-\tan 1\tan (\tan x))(1-\tan 1\tan(\sin x))}\cdot  \frac{\tan(\tan x)-\tan(\sin x)}{x^3} \to \frac12(1+\tan^2 1)$$
with
$$\frac{\tan(\tan x)-\tan(\sin x)}{x^3}=\\=\frac{\tan(\tan x)-\tan x}{x^3}-\frac{\tan(\sin x)-\sin x}{x^3}+\frac{\tan x-\sin x}{x^3} \to \frac13-\frac13+\frac12=\frac12$$
A: I just found a much simpler solution. First we take $\ln$, so we only need to evaluate $$L := \lim_{x\to0}{\frac{\ln{(\tan{(1+\tan x)})} - \ln{(\tan{(1+\sin x)})}}{x^3}}.$$
By Mean Value Theorem, we know that $$\ln{(\tan{(1+\tan x)})} - \ln{(\tan{(1+\sin x)})} = \frac{\sec^2{(1+c)}}{\tan{(1+c)}}(\tan x - \sin x) $$ for some $c \in (\sin x, \tan x)$. Then, using the taylor series of $\tan x = x + \frac{x^3}{3} + \mathcal{O}(x^5)$ and $\sin x = x - \frac{x^3}{6} + \mathcal{O}(x^5)$ immediately gives $$L = \lim_{x \to 0}{\frac{1}{2\sin{(1+c)}\cos{(1+c)}}}$$ where $c \in (\sin x, \tan x)$. Therefore, $$L = \frac{1}{2\sin{1}\cos{1}} = \csc 2.$$ So the original limit is just $e^{\csc 2}$.
A: L'Hospital is very cumbersome for this problem, try this
If $L=\lim_{x\to a}f(x)^{g(x)}\to 1^{\infty}$ (indeterminate), then
$$L=\exp[\lim_{x\to a} [g(x)(f(x)-1)].$$ Here
$$F=\lim_{x\to0}{{\left(\frac{\tan{(1+\tan x)}}{\tan{(1+\sin x)}}\right)}^{1/x^3}}.$$
Then $$F=\exp[\lim_{x\to 0}\frac{1}{x^3}\frac{[\tan(1+\tan x)-\tan(1+\sin x)]}{\tan(1+\sin x)}].$$
Use $\tan A-\tan B= \frac{\sin (A-B)}{\cos A \cos B}$
$$\implies F=\exp[\lim_{x\to 0}\frac{1}{x^3} \frac{\sin(\tan x-\sin x)}{(\cos (1+\tan x)\cos(1+\sin x)\tan(1+\sin x)}].$$
Use $\sin x=x-x^3/6+O(x^5), \tan x=x+x^3/3+O(x^5))$, then
$$\implies F=\exp[\lim_{x\to 0}\frac{1}{x^3}\frac{\sin(x^3/2+O(x^5))}{(\cos (1+\tan x)\cos(1+\sin x)\tan(1+\sin x)}$$
$$\implies F=\exp[\lim_{x\to 0}\frac{1}{x^3}\frac{x^3/2}{(\cos (1+\tan x)\cos(1+\sin x)\tan(1+\sin x)}=\exp[\frac{1}{2\cos^2 1 \tan 1}]$$
Finally, we have $$L=e^{\csc 2}$$
