Answer
$\lim\limits_{x \to 0}\frac{(1+x+sinx)}{3cosx} = \frac{1}{3} = 0.33$
Work Step by Step
$\lim\limits_{x \to 0}\frac{(1+x+sinx)}{3cosx} = \frac{\lim\limits_{x \to 0}(1+x+sinx)}{3\lim\limits_{x \to 0}cosx} = \frac{(1+0+0)}{3(1)} = \frac{1}{3} = 0.33$
