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!
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
$$\lim\limits_{x \to \frac{\pi}{6}^-} f(x) = e^0 = 1.$$
In the "$\log$" expression, expand $\sin$ around $x_0=\frac{\pi}{6}$ using Taylor series, up to the second term, get $$ \sin x \approx \frac{1}{2} + \frac{\sqrt{3}}{2}\left(x-\frac{\pi}{6}\right) $$ so the expression $\log(1-2 \sin x)$ becomes $\log\left(\frac{\sqrt{3}}{2} \left(\frac{\pi}{6} - x\right)\right) = \log \frac{\sqrt{3}}{2} + \log \left(\frac{\pi}{6} - x\right)$. Now set $t=\frac{\pi}{6} - x$, rewrite $-\tan (-t_ = -\frac{\sin t}{\cos t}$ and expand $\sin t \sim t $ for $t \to 0^+$. This additional condition of convergence from the right allows rewriting the limit as
$$ \lim_{t \to 0^{+}} t \log t $$
Now you can rewrite $t \log t = \frac{\log t }{\frac{1}{t}}$, and note that $\frac{1}{t} \to \infty$ and $\log t \to -\infty$. Set $\log t =v, \frac{1}{t} = e^{-v}$ for $v \to \infty$ and obviously $$ \lim_{v \to \infty}\frac{v }{e^v} = 0 $$ All other terms converge to constants and are easy to compute. Keep in mind also that the original expression is $\varphi = e^{\log \varphi}$, so don't forget to take the exponent.
Result: no L'Hopital Rule used, only Taylor Series Expansion!
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 $e^0=1$.
