Answer
$\dfrac{1}{2}$
Work Step by Step
Consider $f(x)=\lim\limits_{x \to 0}\dfrac{\tan x}{x+\sin x}=\lim\limits_{x \to 0} \dfrac{a(x)}{b(x)}$ and $a(0)=0, b(0)=0$
Thus, $f(0)=\dfrac{\pi}{0}$
This shows an Inderminate form of the limit, so apply L-Hospital's rule:
$\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$
Thus,
$\lim\limits_{x \to 0}\dfrac{\sec^2 x}{1+\cos x}=\dfrac{1}{2}$
