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.
