Without using l'hopital's rule 3

Using the small-angle approximation $\sin{x} \approx x$ for $x$ near $0$ ,

$\frac{\sin^2{x}+\sin{2x}}{3x} \approx \frac{x^2+2x}{3x} = \frac{x+2}{3}$

$\lim_{x \rightarrow 0}\frac{\sin^2{x}+\sin{2x}}{3x} = \lim_{x \rightarrow 0}\frac{x+2}{3} = \frac{2}{3} \approx \boxed{0.667}$

$\begin{aligned} L & = \lim_{x \to 0} \frac {\sin^2 x + \sin (2x)}{3x} \\ & = \lim_{x \to 0} \frac {\sin^2 x + 2\sin x \cos x}{3x} \\ & = \lim_{x \to 0} \frac {{\color{#3D99F6}\sin x}(\sin x + 2\cos x)}{3{\color{#3D99F6}x}} & \small \color{#3D99F6} \text{See note.} \\ & = \frac {{\color{#3D99F6}1}(0 + 2(1))}3 \\ & = \frac 23 = \boxed{0.667} \end{aligned}$

Note:

$\begin{aligned} \lim_{x \to 0} \frac {\color{#3D99F6}\sin x}x & = \lim_{x \to 0} \frac {\color{#3D99F6}x-\frac {x^3}{3!}+\frac{x^5}{5!}-...}x & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to 0} \left(1-\frac {x^2}{3!}+\frac{x^4}{5!}-... \right) \\ & = \boxed{1} \end{aligned}$

Converting $\underset { x\rightarrow 0 }{ lim } \frac { { sin }^{ 2 }x+sin(2x) }{ 3x } =\underset { x\rightarrow 0 }{ lim } \frac { { sin }^{ 2 }x+2sinxcosx }{ 3x }$

Using the limit properties we obtain $\frac { 1 }{ 3 } \underset { x\rightarrow 0 }{ lim } \frac { \sin { x } }{ x } \times \underset { x\rightarrow 0 }{ lim } (sinx+2cosx)$

this simplifies to $\frac { 1 }{ 3 } \times 1\times 2=\frac { 2 }{ 3 } \approx 0.67$