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 ?
 A: 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.
A: Choosing the usual branch for the complex logarithm, and keeping $\;z\;$ close to zero in the complex plane, we get
$$\lim_{z\to 0}\frac{\text{Log}\,\frac{\sin z}z}{z^2}\stackrel{\text{l'Hospital}}=\lim_{z\to 0}\frac{\frac{z\cos z-\sin z}{z\sin z}}{2z}=\lim_{z\to 0}\frac{z\cos z-\sin z}{2z^2\sin z}\stackrel{\text{l'H}}=$$
$$=\lim_{z\to 0}\frac{\cos z-z\sin z-\cos z}{4z\sin z+2z^2\cos z}\stackrel{\text{l'H}}=\lim_{z\to 0}\frac{-\sin z-z\cos z}{4\sin z+8z\cos z-2z^2\sin z}\stackrel{\text{l'H}}=$$
$$=\lim_{z\to 0}\frac{-2\cos z+z\sin z}{12\cos z-12z\sin z-2z^2\cos z}=-\frac1{6}$$
Complete the exercise now.
A: 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 $-\frac {\pi}2 \leq x \leq\frac {\pi}2$.
A: $\sin{z}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-......$
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:
$(1-\frac{0^2}{3!}+\frac{0^4}{5!}-......))^{1/z^2}=1^{1/z^2}=\underline{1}$
