<s>[INST] Evaluating $\lim_{x \to \frac{\pi}{6}}{(1-2\sin(x))}^{\tan(\frac{\pi}{6}-x)}$ I'm in a little struggle with this limit, can anyone help me, please?
$$\lim_{x \to \frac{\pi}{6}}{(1-2\sin(x))}^{\tan(\frac{\pi}{6}-x)}$$
I tried to use the logarithm to then use L'Hospital's rule but I got stuck here: 
$\ln(L)=\lim_{x \to \frac{\pi}{6}}{[\tan(\frac{\pi}{6}-x)\ln(1-2\sin(x))]}$
Thank you!
 [/INST] Let $f(x) = (1-2\sin x)^{\tan(\frac{\pi}{6}-x)}$, then $f(x) = e^{g(x)}$ with $g(x) = \tan(\frac{\pi}{6}-x) \log (1-2\sin x)$.
$$\begin{align}
\lim\limits_{x \to \frac{\pi}{6}^- } g(x)
&= \lim\limits_{x \to \frac{\pi}{6}^- } \frac{\tan\left(\frac{\pi}{6}-x\right)}{\frac{\pi}{6}-x} \left(\frac{\pi}{6}-x\right)\log \left(1-2\sin x\right)
\\
&\overset{(1)}{=} \lim\limits_{x \to \frac{\pi}{6}^- } \left(\frac{\pi}{6}-x\right)\log \left(1-2\sin x\right)
\\
&=\lim\limits_{x \to \frac{\pi}{6}^- } \frac{ \log (1-2\sin x)}{\frac{1}{\frac{\pi}{6}-x}}
\\
&\overset{\mathrm{H}}{=} \lim\limits_{x \to \frac{\pi}{6}^-} (-2\cos x)\frac{\left(\frac{\pi}{6}-x\right)^2}{1-2\sin x}
\\
&= -\sqrt{3}\lim\limits_{x \to \frac{\pi}{6}^-} \frac{\left(\frac{\pi}{6}-x\right)^2}{1-2\sin x}
\\
&\overset{\mathrm{H}}{=}
-\sqrt{3}\lim\limits_{x \to \frac{\pi}{6}^-}\frac{-2\left(\frac{\pi}{6}-x\right)}{-2\cos x }
\\
&= 0
\end{align}$$
where in $(1)$ I have used $\lim_{y\to0} \frac{\tan y}{y} = 1$ and $H$ denotes the usage of L'Hôpital's rule. 
Hence, we conclude that 
$$\lim\limits_{x \to \frac{\pi}{6}^-} f(x) = e^0 = 1.$$
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