TITLE: Evaluating $\lim_{x \to \frac{\pi}{6}}{(1-2\sin(x))}^{\tan(\frac{\pi}{6}-x)}$
QUESTION [5 upvotes]: 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!

REPLY [2 votes]: Your job might be simpler if you substitute $\pi/6-x=2t$. Then
$$
1-2\sin x=1-2\sin(\pi/6-2t)=1-\cos 2t+\sqrt{3}\sin 2t=2\sin t(\sin t+\sqrt{3}\cos t)
$$
Note that in order that the limit makes sense you need $\sin x<1/2$, so $0<x<\pi/6$ (the lower bound is mostly irrelevant, though), hence $t>0$.
How does this help? You get to evaluate the limit for $t\to0$ of
$$
\tan2t\bigl(\log(\sin t)+\log(2\sin t+2\sqrt{3}\cos t)\bigr)
$$
The part $\tan2t\log(2\sin t+2\sqrt{3}\cos t)$ poses no problem: its limit is $0$. Then you need to compute the limit of
$$
\frac{2\cos t}{\cos2t}\sin t\log\sin t
$$
The fraction part has limit $2$. The part $\sin t\log\sin t$ has limit $0$, as it's easy to show with l'Hôpital or other methods.
Hence the limit is $0$. Therefore your original limit is $e^0=1$.