calculus - Compute $lim_{thetarightarrow 0}frac{sin{(tan{theta})}-sin{(sin{theta})}}{tan{(tan{theta})}-tan{(sin{theta})}}$

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 $$\lim_{\theta\rightarrow0}\frac{\sin{(\tan{\theta})-\sin{(\sin{\theta})}}}{\sin{(\tan{\theta}-\sin{\theta})}}.$$

Answer

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 $\tan(\theta)$.