Pieces of a Limit

$\lim _{ x\rightarrow 0 }{ \left( \frac { \sin { 2x } }{ { x }^{ 3 } } +a+\frac { b }{ { x }^{ 2 } } \right) } =a+\lim _{ x\rightarrow 0 }{ \left( \frac { \sin { 2x } +bx }{ { x }^{ 3 } } \right) } =a+\lim _{ x\rightarrow 0 }{ \left( \frac { 2\cos { 2x } +b }{ 3{ x }^{ 2 } } \right) } =a+\left( b+2 \right) \infty =0$

So, $b+2=0\Rightarrow b=-2$

Then, $a+\lim _{ x\rightarrow 0 }{ \left( \frac { 2\cos { 2x } -2 }{ 3{ x }^{ 2 } } \right) } =a+\lim _{ x\rightarrow 0 }{ \left( \frac { -4\sin { 2x } }{ 6x } \right) } =a+\frac { -4 }{ 3 } =0\Rightarrow a=\frac { 4 }{ 3 }$

To conclude, $3a+b=3·\frac { 4 }{ 3 } +\left( -2 \right) =2$

This problem can be done by another method i.e by expanding $sin(2x)$ and consider up to second place then after manipulating expression it would give hence $b=-2$ and $a=$ $\frac{4}{3}$ .Therefore answer is 2.