A Little Twist in Kinshuk Singh's Problem

${ lim }_{ n\rightarrow \infty }\left| \sin { \left( 3\pi \sqrt { { n }^{ 4 }+{ n }^{ 2 } } \right) } \right| ={ lim }_{ n\rightarrow \infty }\left| \sin { \left( 3\pi { n }^{ 2 }{ \left( 1+\cfrac { 1 }{ { n }^{ 2 } } \right) }^{ 1/2 } \right) } \right| \\ \\ \because { n }^{ 2 }{ \left( 1+\cfrac { 1 }{ { n }^{ 2 } } \right) }^{ 1/2 }={ n }^{ 2 }{ \left( 1+\cfrac { 1 }{ { 2n }^{ 2 } } -\cfrac { 1 }{ 8{ n }^{ 4 } } +... \right) }={ n }^{ 2 }+\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 8{ n }^{ 2 } } +...\\ \\ \therefore { lim }_{ n\rightarrow \infty }\left| \sin { \left( 3\pi \sqrt { { n }^{ 4 }+{ n }^{ 2 } } \right) } \right| ={ lim }_{ n\rightarrow \infty }\left| \sin { \left( 3\pi { \left( { n }^{ 2 }+\cfrac { 1 }{ 2 } -0+... \right) } \right) } \right| \\ \\ ={ lim }_{ n\rightarrow \infty }\left| \sin { \left( 3{ n }^{ 2 }\pi +\cfrac { 3\pi }{ 2 } \right) } \right| =\left| \pm \sin { \left( \cfrac { 3\pi }{ 2 } \right) } \right| =1$