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.
 A: $$\lim _{ x\rightarrow 0 }{ { \left( \frac { \sin { x }  }{ x }  \right)  }^{ \frac { 1 }{ { x }^{ 2 } }  } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { \sin { x }  }{ x } -1 \right)  }^{ \frac { 1 }{ { x }^{ 2 } }  } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { \sin { x-x }  }{ x }  \right)  }^{ \frac { 1 }{ { x }^{ 2 } }  } } =$$ 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 $$\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \cos { x } -1 }{ 3{ x }^{ 2 } }  }  }=\\ \overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\sin { x }  }{ 6{ x } }  }  }=\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\cos { x }  }{ 6 }  }  }={ e }^{ -\frac { 1 }{ 6 }  }$$
A: You may just use $\lim_{z\to +\infty}\left(1+\frac{\alpha}{z}\right)^z = e^{\alpha}$ and squeezing, since in a neighbourhood of the origin
$$ 1-\frac{x^2}{6}\leq \frac{\sin x}{x} \leq e^{-x^2/6} $$
holds.
A: hint:use this fact $$\lim_{x \to 0 }(1+x)^{\dfrac1x}=e$$
$$\lim_{x \rightarrow 0}(\frac{sin x}{x})^{\dfrac{1}{x^2}}=\\
\lim_{x \rightarrow 0}(\dfrac{x-\dfrac{x^3}{6}+o(x^5)}{x})^{\dfrac{1}{x^2}}=\\
\lim_{x \rightarrow 0}(1-\dfrac{x^2}{6})^{\dfrac{1}{x^2}}=\\
\lim_{x \rightarrow 0}(1-\dfrac{x^2}{6})^{\dfrac{-6}{x^2}\times \dfrac{1}{-6}}=\\
\lim_{x \rightarrow 0}((1-\dfrac{x^2}{6})^{\dfrac{-6}{x^2})^{ \dfrac{1}{-6}}}=\\
e^{\dfrac{1}{-6}}$$
A: Hint:
Rewrite the limit as
$$\lim_{x\to 0}\exp\left( \frac{1}{x^2} \ln \left(\frac{\sin x}{x} \right) \right)= \exp \left(\lim_{x\to 0} \frac{1}{x^2} \ln \left(\frac{\sin x}{x}\right) \right)$$
and apply L'Hospital rule on
$$\lim_{x\to 0} \frac{1}{x^2} \ln \left(\frac{\sin x}{x}\right).$$
This is possible since the exponential function $\exp$ is continuous.
You should also know that $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\ln 1=0$.
A: $$
\begin{aligned}
\lim _{x\to 0}\left(\frac{\sin \left(x\right)}{x}\right)^{\frac{1}{x^2}}\:
& = e^{\lim _{x\to 0}\left(\frac{1}{x^2}\cdot \:ln\left(\frac{sinx}{x}\right)\right)}
\\& = e^{\lim _{x\to 0}\left(\frac{1}{x^2}\cdot \ln\left(\frac{x-\frac{x^3}{6}+o\left(x^3\right)}{x}\right)\right)}
\\& = e^{\lim \:_{x\to \:0}\left(\frac{1}{x^2}\cdot \ln\left(1-\frac{x^2}{6}+o\left(x^3\right)\right)\right)}
\\& \approx_0 e^{\lim \:_{x\to \:0}\left(-\frac{x^2}{x^26}\right)}
\\& = \color{red}{e^{-\frac{1}{6}}}
\end{aligned}$$
Solved with Taylor expansion $(\sin(x) = x-\frac{x^3}{6}+o(x^3))$ and asymptotic approximations $(\ln(1+x) \approx_0 x)$
