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!
Best Answer
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
$$\lim\limits_{x \to \frac{\pi}{6}^-} f(x) = e^0 = 1.$$