<s>[INST] 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 ?
 [/INST] $\sin{z}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-......$
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:
$(1-\frac{0^2}{3!}+\frac{0^4}{5!}-......))^{1/z^2}=1^{1/z^2}=\underline{1}$
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