#### Answer

$\dfrac{1}{3}$

#### Work Step by Step

Note that, $\lim\limits_{x \to a}\sin x=\sin a$ and $\lim\limits_{x \to a}\cos x=\cos a$ for all real numbers $a$. Using this along with our limit laws, we get
$\lim\limits_{x \to 0}\dfrac{1+x+\sin x}{3\cos x}=\dfrac{1+0+\sin 0}{3\cos 0}=\dfrac{1+0+0}{3(1)}=\dfrac{1}{3}.$