How to evaluate $\lim\limits_{x \to 0^{-}} (1-8^x)^{\sin 3x}$? How to evaluate 
$$
\lim\limits_{x \to 0^{-}} (1-8^x)^{\sin 3x}
$$
Does someone have at least a hint on how to start? I'm clueless.
 A: By Taylor series
$$(1-8^x)^{\sin(3x)}=\exp(\sin(3x)(\ln(1-\exp(x\ln8))))\sim\exp(3x\ln(-x\ln8))\xrightarrow{x\to 0^-}1$$
A: It is best to take logs. Let $L$ be the desired limit and then we have
\begin{align}
\log L &= \log\left(\lim_{x \to 0^{-}}(1 - 8^{x})^{\sin 3x}\right)\notag\\
&= \lim_{x \to 0^{-}}\log(1 - 8^{x})^{\sin 3x}\text{ (by continuity of log)}\notag\\
&= \lim_{x \to 0^{-}}\sin 3x \log(1 - 8^{x})\notag\\
&= \lim_{x \to 0^{-}}\frac{\sin 3x}{3x}\cdot 3x\log(1 - 8^{x})\notag\\
&= 3\lim_{x \to 0^{-}}\frac{\sin 3x}{3x}\cdot\lim_{x \to 0^{-}}x\log(1 - 8^{x})\tag{1}\\
&= 3\lim_{y \to 0^{+}}\frac{\log(1 - y)}{\log 8}\log y\text{ (putting }y = 1 - 8^{x})\notag\\
&= \frac{3}{\log 8}\lim_{y \to 0^{+}}\frac{\log (1 - y)}{-y}\cdot(-y)\log y\tag{2}\\
&= -\frac{3}{\log 8}\lim_{y \to 0^{+}}1\cdot y\log y\tag{3}\\
&= -\frac{3}{\log 8}\cdot 0 = 0
\end{align}
so that $L = 1$. We have used the following standard limits
\begin{align}
\lim_{z \to 0}\frac{\sin z}{z} &= 1\text{ (used in (1) by putting }z = 3x)\notag\\
\lim_{z \to 0}\frac{\log(1 + z)}{z} &= 1\text{ (used in (2) by putting }z = -y)\notag\\
\lim_{y \to 0^{+}}y\log y & = 0\text{ (used in (3))}\notag
\end{align}
