calculus - Evaluate $lim_{xrightarrow 0} frac{sin x}{x + tan x}$ without L'Hopital

I need help finding the the following limit:

$$\lim_{x\rightarrow 0} \frac{\sin x}{x + \tan x}$$

I tried to simplify to:

$$\lim_{x\rightarrow 0} \frac{\sin x \cos x}{x\cos x+\sin x}$$

but I don't know where to go from there. I think, at some point, you have to use the fact that $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$. Any help would be appreciated.

Thanks!

Answer

$$\frac{\sin x}{x + \tan x} = \frac{1}{\frac{x}{\sin x}+\frac{\tan x}{\sin x}} \to 1/2$$