TITLE: Limit of $\lim\limits_{x \to 0} \Bigl(\frac{\sin{x}}{x}\Bigr)^{1/x^{3}}$
QUESTION [6 upvotes]: How to evaluate the limit: $$\lim_{x \to 0} \Bigl(\frac{\sin{x}}{x}\Bigr)^{1/x^{3}}$$
I think it goes to $1$ because $\lim\limits_{x \to 0} \frac{\sin{x}}{x} =1$ and so power of $1$ should also be $1$. Am I right?

REPLY [2 votes]: Hint: If allowed, I would use the Maclaurin series for $\sin(x)=x+\frac{x^3}{6}+O(x^5)$. Then, after diving by $x$, note that $\lim\limits_{x\to0}(1+x^2)^{1/x^3}=\lim\limits_{x\to0}(1+x^2)^{(1/x^2)(1/x)}=\lim\limits_{x\to0}e^{1/x}$.