How do you find this limit with a relationship to $e$ using Taylor series? 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 $x\to 0$.
