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 $$\lim_{\theta\rightarrow0}\frac{\sin{(\tan{\theta})-\sin{(\sin{\theta})}}}{\sin{(\tan{\theta}-\sin{\theta})}}.$$
 A: $$\begin{align}&\lim_{\theta\rightarrow0}\frac{\sin{(\tan{\theta})-\sin{(\sin{\theta})}}}{\sin{(\tan{\theta}-\sin{\theta})}} &\\=& \lim_{\theta\rightarrow0}\frac{2 \sin\left({\dfrac{\tan{\theta} -\sin{\theta}}2}\right)\cos\left({\dfrac{\tan{\theta} +\sin{\theta}}2 } \right)}{2\sin{\left(\dfrac{\tan{\theta}-\sin{\theta}}{2}\right)}\cos{\left(\dfrac{\tan{\theta}-\sin{\theta}}{2}\right)}}&\\ =&\lim_{\theta\rightarrow0}\frac{\cos\left({\dfrac{\tan{\theta} +\sin{\theta}}2 } \right)}{\cos{\left(\dfrac{\tan{\theta}-\sin{\theta}}{2}\right)}} = 1\end{align}$$
A: Using the Mean Value Theorem,
$$
\begin{align}
\lim_{\theta\to0}\frac{\sin(\tan(\theta))-\sin(\sin(\theta))}{\tan(\tan(\theta))-\tan(\sin(\theta))}
&=\lim_{\theta\to0}\frac{\frac{\sin(\tan(\theta))-\sin(\sin(\theta))}{\tan(\theta)-\sin(\theta)}}{\frac{\tan(\tan(\theta))-\tan(\sin(\theta))}{\tan(\theta)-\sin(\theta)}}\\
&=\lim_{\theta\to0}\frac{\cos(\xi_1(\theta))}{\sec^2(\xi_2(\theta))}\\[6pt]
&=\frac{\cos(0)}{\sec^2(0)}\\[12pt]
&=1
\end{align}
$$
where $\xi_1(\theta)$ and $\xi_2(\theta)$ are between $\sin(\theta)$ and $\tan(\theta)$.
A: 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 $\cos 0 = 1$
A: \begin{align}
\lim_{\theta\rightarrow 0}\frac{\sin{(\tan{\theta})-\sin{(\sin{\theta})}}}{\sin{(\tan{\theta}-\sin{\theta})}} &= \lim_{\theta\rightarrow 0}\frac{\sin(\tan{\theta})-\sin(\sin{\theta})}{{\frac{\sin(\tan{\theta}-\sin{\theta})}{\tan\theta - \sin\theta}}(\tan\theta - \sin\theta)} \\
&= \lim_{\theta\rightarrow 0}\frac{2\cos(\frac{\tan \theta +\sin\theta}{2})\sin(\frac{\tan \theta -\sin\theta}{2})}{(\tan\theta - \sin\theta)}\\
&=1
\end{align}
