$=\frac{1}{6}\cdot \lim _{x\to 0}(\frac{\sin(x)+\sin(5x)}{x})$
Apply L'Hopital's Rule$=\frac{1}{6}\cdot \lim _{x\to 0}(\frac{\cos(x)+\cos(5x)\cdot 5}{1})$
Plug in the value $ x=0$$=\frac{1}{6}\cdot \frac{\cos(0)+\cos(5\cdot 0)\cdot 5}{1}$
Simplify $\frac{1}{6}\cdot \frac{\cos(0)+\cos(5\cdot 0)\cdot 5}{1}:{\quad}1$$=1$