Limited yet powerful (VII)

See below for the step-by-step solution:

$lim_{x\rightarrow0} (\frac {x}{sin {x}})^{\frac {1}{x^2}}$ = $e^{ln \cdot lim_{x\rightarrow0} (\frac {x}{sin {x}})^{\frac {1}{x^2}}}$ . If the last limit exists, considerating that the function f(x)=ln x is continuous we can do $e^{ln \cdot lim_{x\rightarrow0} (\frac {x}{sin {x}})^{\frac {1}{x^2}}}$ = $e^{lim_{x\rightarrow0} ln (\frac {x}{sin {x}})^{\frac {1}{x^2}}}$ . Now, we are going to prove that the last limit exists and this will imply that we can do it.

$lim_{x\rightarrow0} ln (\frac {x}{sin {x}})^{\frac {1}{x^2}}$ = $lim_{x\rightarrow0} \frac {1}{x^2} \cdot ln (\frac {x}{sin {x}})$ . Applying L'Hopital

$lim_{x\rightarrow0} \frac {1}{x^2} \cdot ln (\frac {x}{sin {x}})$ = $lim_{x\rightarrow0} \frac {sin (x)}{x} \cdot (\frac {sin (x) - x \cdot cos(x)}{sin^{2}{x} \cdot 2x})$ = $lim_{x\rightarrow0} \frac {sin (x) - x \cdot cos(x)}{sin(x) \cdot 2 \cdot x^2}$ = $lim_{x\rightarrow0} \frac {(x - \frac{x^3}{3!} + o(x^3)) - (x - \frac {x^3}{2!} + o(x^3))}{ 2 \cdot x^3}$ = $\frac {1}{6}$ ...