Answer
$$\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{x + \sin x}} \cr
& {\text{Evaluating, we get:}} \cr
& = \frac{{\tan 0}}{{0 + \sin 0}} \cr
& = \frac{0}{0} \cr
& {\text{Using l'Hopital's rule:}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\tan x} \right)}}{{\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {x + \sin x} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sec }^2}x}}{{1 + \cos x}} \cr
& {\text{Evaluating the limit, we get:}} \cr
& = \frac{{{{\sec }^2}0}}{{1 + \cos 0}} \cr
& = \frac{1}{{1 + 1}} \cr
& = \frac{1}{2} \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{x + \sin x}} = \frac{1}{2} \cr} $$
