Answer
$$\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta+\tan \theta}=\frac{1}{2}
$$
Work Step by Step
Given $$\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta+\tan \theta}$$
by dividing numerator and denominator by $\theta$, we get
\begin{align}
\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta+\tan \theta}&=\lim _{\theta \rightarrow 0} \frac{\frac{\sin \theta}{\theta}}{1+\frac{\sin \theta}{\theta} \cdot \frac{1}{\cos \theta}}\\
&=\frac{\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}}{1+\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} \lim _{\theta \rightarrow 0} \frac{1}{\cos \theta}}\\
&=\frac{1}{1+1 \cdot 1}=\frac{1}{2}, \ \ \ (since \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1 )
\end{align}