show that $\lim_{x\rightarrow 0} \left(\frac{\sin(x)}{x}\right)^{1/x^2} = e^{-1/6}$ I see that this is indeterminate form, so I approach L'Hopital rule, but I can not find this limit. Please help me to find this limit.

$$\lim _{ x\rightarrow 0 }{ { \left( \frac { \sin { x }  }{ x }  \right)  }^{ \frac { 1 }{ { x }^{ 2 } }  } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { \sin { x }  }{ x } -1 \right)  }^{ \frac { 1 }{ { x }^{ 2 } }  } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { \sin { x-x }  }{ x }  \right)  }^{ \frac { 1 }{ { x }^{ 2 } }  } } =$$ hence $\lim _{ x\rightarrow 0 }{ \frac { \sin { x-1 }  }{ x }  } =0\quad $ we can use here a well known limit form $$\lim _{ x\rightarrow 0 }{ { \left( 1+x \right)  }^{ \frac { 1 }{ x }  } } =e$$
then we can write it as $$=\lim _{ x\rightarrow 0 }{ { { \left[ { \left( 1+\frac { \sin { x } -x }{ x }  \right)  }^{ \frac { x }{ \sin { x } -x }  } \right]  }^{ \frac { \sin { x } -x }{ x } \cdot \frac { 1 }{ { x }^{ 2 } }  } } } $$
the inner part of limit is clearly is $e$ so $$=\quad { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \sin { x } -x }{ x } \cdot \frac { 1 }{ { x }^{ 2 } }  }  }$$
now to find the exponent limit we use L'Hospital rule here three times $$\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \cos { x } -1 }{ 3{ x }^{ 2 } }  }  }=\\ \overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\sin { x }  }{ 6{ x } }  }  }=\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\cos { x }  }{ 6 }  }  }={ e }^{ -\frac { 1 }{ 6 }  }$$
