Taylor Series to the Power 1/z

I am attempting to find the Taylor Series for $(\frac{\sin{z}}{z})^{\frac{1}{z^2}}$. While I can plug this into Wolfram and use the output, I want to understand how to calculate the Taylor Series myself. I can find the Taylor Series for $\frac{\sin{z}}{z}$ easily enough, but I cannot figure out for the power $\frac{1}{z^2}$ plays into things.

I just need to find the first few terms so that I can evaluate the limit, I do not need the nth term.

Solution 1:

Write $$ \Bigl(\frac{\sin z}{z}\Bigr)^{1/z^2}=\exp\Bigl(\frac{1}{z^2}\,\log\Bigl(\frac{\sin z}{z}\Bigr)\Bigr) $$ I would first find a few terms of $$ \frac{1}{z^2}\,\log\Bigl(\frac{\sin z}{z}\Bigr)=\frac{1}{z^2}\,\log\Bigl(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\dots\Bigr) $$ using $$ \log(1+w)=w-\frac{w^2}{2}+\frac{w^3}{3}-\frac{w^4}{4}+\dots $$ and then use the Taylor expansion of $e^w$.