Compute $\lim_{\theta\rightarrow 0}\frac{\sin{(\tan{\theta})}-\sin{(\sin{\theta})}}{\tan{(\tan{\theta})}-\tan{(\sin{\theta})}}$ I rewrote it by writing the tan as sin/cos and cross multiplying:
$$\frac{\sin{(\tan{\theta})}-\sin{(\sin{\theta})}}{\tan{(\tan{\theta})}-\tan{(\sin{\theta})}}= \frac{\sin{(\tan{\theta})}-\sin{(\sin{\theta})}}{\frac{\sin{(\tan{\theta})}\cos{(\sin{\theta})-\cos{(\tan{\theta})\sin{(\sin{\theta})}}}}{\cos{(\tan{\theta})\cos{(\sin{\theta})}}}}.$$
Using addition formula for sine i get: $$\sin{(\tan{\theta})}\cos{(\sin{\theta})-\cos{(\tan{\theta})\sin{(\sin{\theta})}}}=\sin{(\tan{\theta}}-\sin{\theta}), $$
Since $\cos{(\tan{\theta})}\cos{(\sin{\theta})}\rightarrow1$ as $\theta\rightarrow 0,$ the problem is reduced to finding the limit of $$\lim_{\theta\rightarrow0}\frac{\sin{(\tan{\theta})-\sin{(\sin{\theta})}}}{\sin{(\tan{\theta}-\sin{\theta})}}.$$

Let $x = \sin \theta$. Then
$$
\lim_{\theta\rightarrow 0}\frac{\sin{(\tan{\theta})}-\sin{(\sin{\theta})}}{\tan{(\tan{\theta})}-\tan{(\sin{\theta})}}\\
= \lim_{x\rightarrow 0}\frac{\sin{(\frac{x}{\sqrt{1 - x^2}})}-\sin{(x)}}{\tan{(\frac{x}{\sqrt{1 - x^2}})}-\tan{(x)}}\\
$$
Further, by trig theorems (e.g. in wikipedia) , 
$\sin a - \sin b = 2 \cos ((a+b)/2) \sin ((a-b)/2)$  
and 
$\tan a - \tan b =  \frac{\sin (a-b) }{\cos a \cos b} = \frac{2 \sin ((a-b)/2)\cos ((a-b)/2)  }{\cos a \cos b} = $
So we get
$\frac{\sin a - \sin b}{\tan a - \tan b} = \frac{ \cos ((a+b)/2) \cos a \cos b}{\cos ((a-b)/2)  }
$
So we have
$$
 \lim_{x\rightarrow 0}\frac{\cos{((\frac{x}{\sqrt{1 - x^2}} + x)/2)}\cos{(x)}\cos{(\frac{x}{\sqrt{1 - x^2}})}}{\cos{((\frac{x}{\sqrt{1 - x^2}} - x)/2)}} = 1
$$
since all angular functions  give $\cos 0 = 1$
