Answer
$$\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta+\tan\theta}=\frac{1}{2}$$
Work Step by Step
$$A=\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta+\tan\theta}$$ $$A=\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta+\frac{\sin\theta}{\cos\theta}}$$ Divide both numerator and denominator by $\theta$ $$A=\lim\limits_{\theta\to0}\frac{\frac{\sin\theta}{\theta}}{1+\frac{\sin\theta}{\theta\cos\theta}}$$ $$A=\frac{\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta}}{{\lim\limits_{\theta\to0}}(1+\frac{\sin\theta}{\theta\cos\theta})}$$ $$A=\frac{1}{1+\lim\limits_{\theta\to0}(\frac{\sin\theta}{\theta}\times\frac{1}{\cos\theta})}$$ $$A=\frac{1}{1+(1\times\frac{1}{\cos0})}$$ $$A=\frac{1}{1+(1\times1)}$$ $$A=\frac{1}{2}$$
