find this limit $\lim_{x\to0^{+}}\frac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$ find the limit.

$$\lim_{x\to0^{+}}\dfrac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$$

my try:
$$\tan{x}=x+\dfrac{1}{3}x^3+o(x^3),\sin{x}=x-\dfrac{1}{6}x^3+o(x^3)$$
so
$$\tan{(\tan{x})}=\tan{x}+\dfrac{1}{3}(\tan{x})^3+o(\tan^3{x})=x+\dfrac{1}{3}x^3+\dfrac{1}{3}(x+\dfrac{1}{3}x^3)^3+o(x^3)$$
$$\tan{(\sin{x})}=\sin{x}+\dfrac{1}{3}(\sin{x})^3+o(\sin^3{x})=x-\dfrac{1}{6}x^3+\dfrac{1}{3}(x-\dfrac{1}{6}x^3)^3+o(x^3)$$
so
$$\tan{(\tan{x})}-\tan{(\sin{x})}=\dfrac{1}{2}x^3+o(x^3)$$
$$\tan{x}-\sin{x}=\dfrac{1}{2}x^3+o(x^3)$$
Have other metods? Thank you 
 A: I think this identity is useful here:
$$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$$ and the fact that $\tan(a)\sim a$ while $a\sim 0$.
A: Yours is easily the simplest $direct$ method.
My first idea for the limit is that that really looks like a difference quotient for the derivative of $\tan$. Distilling the core idea, I make the following conjecture:

Let $f(x)$ be continuously differentiable at $a$. Then,
  $$ \lim_{\substack{(x,y) \to (a,a) \\ x \neq y}} \frac{f(x) - f(y)}{x-y} = f'(a)$$

which lets us immediately see the limit should be
$$\tan'(0) = \sec^2(0) = 1$$
A: Since all the functions involved are continuosly differentiable we can apply the Mean Value Theorem to get
$$
\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}=\frac1{\cos^2 z}
$$
for some $z(x)$ with $\sin x\leq z(x)\leq\tan x$. By the Squeeze Rule, $z\to0$ when $x\to0$, and so the limit is $1/\cos^20=1$.
A: Factor a $\tan(x)$ from the denominator:
$$\frac{\tan\tan(x) - \tan\sin(x)}{\tan(x)-\sin(x)} = \frac{\frac{\tan\tan(x)}{\tan(x)} - \cos(x)\frac{\tan\sin(x)}{\sin(x)}}{1-\cos(x)}.$$
The quotients in the numerator go to $1$, so the numerator goes to $1-\cos(x)$. This suggests cancellation.
After two derivatives, the denominator of this quotient will go to $1$, so one strategy is L'Hopital's rule. (I don't have time to compute it but maybe I'll have time later today.)
