The limit in question is
$$ \lim_{x \to 0}\left(\frac{\sin(x)}{x}\right)^{1/x^2} $$
When I replace $\sin(x)$ with its Taylor series about $0$ and cancel out the $x$, I get
$$ \lim_{x \to 0}\left(1-\frac{x^2}{6} + \frac{x^4}{5!} \mp \cdots\right)^{1/x^2} $$
The answer in the book is $e^\frac{-1}{6}$. If I only look at the $x^2$ term, I can see where $-1/6$ comes from. I'm just not sure how I can definitively say the answer is $e^\frac{-1}{6}$. Why can I discount the other powers of $x$?
To handle the higher power terms, you can use the big O notation.
Note that $$ \frac{\sin x}{x}=\frac1x\left(x-\frac{x^3}{6}+O(x^5)\right)=1-\frac{x^2}{6}+O(x^4) $$ and $$ \ln(1-y)=-y-\frac{y^2}{2}-\frac{y^3}{3}+O(y^5) $$ So $$ \begin{align} \left(\frac{\sin x}{x}\right)^{1/x^2} =&\exp\left(\frac{1}{x^2}\ln(\frac{\sin x}{x})\right)\\ =&\exp\left(\frac{1}{x^2}\ln(1-\frac{x^2}{6}+O(x^4))\right)\\ =&\exp\left(\frac{1}{x^2}(-\frac{x^2}{6}+O(x^4))\right)\\ =&\exp\left(-\frac{1}{6}+O(x^2)\right)\to e^{-1/6} \end{align} $$ as $x\to 0$.