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

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 $a\sim 0$.

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

$$\tan'(0) = \sec^2(0) = 1$$

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 $1/\cos^20=1$.