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 ?

You were in the good track. Starting with $$A=\left(\dfrac{\sin z}{z}\right)^{1/z^2}$$ $$\log(A)=\frac{1}{z^2}\log\left(\frac{\sin z}z\right)$$ Now, as Ian did, use the Taylor expansion $$\sin(z)=z-\frac{z^3}{6}+\frac{z^5}{120}-\frac{z^7}{5040}+O\left(z^8\right)$$ $$\frac{\sin z}z=1-\frac{z^2}{6}+\frac{z^4}{120}-\frac{z^6}{5040}+O\left(z^7\right)$$ Now, consider $$\log(1-y)=-y-\frac{y^2}{2}-\frac{y^3}{3}+O\left(y^4\right)$$ and use $$y=\frac{z^2}{6}-\frac{z^4}{120}+\frac{z^6}{5040}$$ and arrive to $$\log\left(\frac{\sin z}z\right)=-\frac{z^2}{6}-\frac{z^4}{180}-\frac{z^6}{2835}+O\left(z^8\right)$$ $$\log(A)=-\frac{1}{6}-\frac{z^2}{180}-\frac{z^4}{2835}+O\left(z^6\right)$$ that is to say $$A=e^{-\frac{1}{6}-\frac{z^2}{180}-\frac{z^4}{2835}+\cdots}=e^{-\frac{1}{6}}e^{-\frac{z^2}{30}-\frac{2z^4}{945}}$$ Use that $$e^y=1+y+\frac{y^2}{2}+O\left(y^3\right)$$ and replace $$y=-\frac{z^2}{30}-\frac{2z^4}{945}$$ and you should finally arrive to $$A=e^{-\frac{1}{6}}\Big(1-\frac{z^2}{180}-\frac{17 z^4}{50400}+\cdots\Big)$$ which also shows how the limit is approached.
If you plot the original function and the last approximation on the same graph, you will probably be amazed to see how close are the curves for $-\frac {\pi}2 \leq x \leq\frac {\pi}2$.
