Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 57: 47

$\dfrac{1}{3}$

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 $\lim\limits_{x \to 0}\dfrac{1+x+\sin x}{3\cos x}=\dfrac{1+0+\sin 0}{3\cos 0}=\dfrac{1+0+0}{3(1)}=\dfrac{1}{3}.$