Value of $\lim\limits_{z \to 0}\bigl(\frac{\sin z}{z}\bigr)^{1/z^2}$ Find the value of $$\lim\limits_{z \to 0}\left(\dfrac{\sin z}{z}\right)^{1/z^2}$$

So I took a log: $$\frac{1}{z^2}\log\left(\frac{\sin z}z\right)$$ If I could expand it something like $\log(1+x)$ .. any hints ?
Using the Maclaurin series for $\sin$, we have

$$\frac{\sin(z)}{z} = \frac{z-\frac{z^3}{6}+o \left ( z^3 \right )}{z} = 1-\frac{z^2}{6}+o\left ( z^2 \right ).$$

Using the most common Taylor series for $\ln$, we have 

$$\ln(1+z)=z+o(z).$$

Combining these we get

$$\frac{1}{z^2} \ln \left ( \frac{\sin(z)}{z} \right ) = \frac{1}{z^2} \left ( -\frac{z^2}{6} + o(z^2) \right ) = -\frac{1}{6} + o(1).$$

I think you can take it from here. An approach without using series would have been to use L'Hopital's rule a couple times, though this would have involved some ugly algebra.
$\sin{z}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-......$

Therefore, we have

$\lim_{z \to 0}({\frac{\sin{z}}{z}})^{1/z^2}=\lim_{z \to 0}(\frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}-......}{Z})^{1/z^2}$

=$\lim_{z \to 0}(1-\frac{z^2}{3!}+\frac{z^4}{5!}-......)^{1/z^2}$

=$(\lim_{z \to 0}(1-\frac{z^2}{3!}+\frac{z^4}{5!}-......))^{1/z^2}$ (the limit of expression inside the bracket is taken)

This gives:

$(1-\frac{0^2}{3!}+\frac{0^4}{5!}-......))^{1/z^2}=1^{1/z^2}=\underline{1}$