Help! Totally stuck up with this limit.. I don't even know where to start with this
$$\displaystyle\lim_{x \rightarrow \frac{\pi}{6}} (2+\cos {6x})^{\ln |\sin {6x}|}$$
Please help me out (Hints in the right direction would be appreciated)
 A: When $f(x)\to1$ and $g(x)\to\infty$ then in order to find $\lim f(x)^{g(x)}$, first find $\log\lim f(x)^{g(x)} = \lim\log(f(x)^{g(x)})= \lim(g(x)\log f(x))=L$ and conclude that $\lim f(x)^{g(x)}=\exp L$.
In this case you I'd try L'Hopital's rule applied to $\dfrac{\log f(x)}{1/g(x)}$.
A: Let $t=\frac\pi6-x$ then $\sin(6x)=\sin(6u)$ and $\cos(6x)=-\cos(6u)$ so the desired limit becomes
$$\lim_{u\to0}(1+(1-\cos(6u))^{\ln|\sin(6u)|}\sim_{u\to0}\exp(\ln|6u|(1-\cos(6u)))\sim_0\exp(\ln|6u|18u^2)\sim_01$$
A: $$\lim_{x \rightarrow \frac{\pi}{6}} (2+\cos {6x})^{\ln |\sin {6x}|}=
\lim_{t \rightarrow 0} (2+(-1))^{\ln |t|}=\lim_{t \rightarrow 0} 1^{\ln |t|}=\lim_{t \rightarrow 0} 1=1$$
EDIT: The procedure above is not correct because it has $1^\infty$ indeterminate step.
I will follow the correct procedure as shown in Michael Hardy's answer.
Let $x=\pi/6-t/6$, then the limit becomes
$$\lim_{t \rightarrow 0} (2-\cos {t})^{\ln |\sin {t}|}$$
We first consider
$$\lim_{t \rightarrow 0^{+}} (2-\cos t)^{\ln \sin t}=:\lim_{t \rightarrow 0^{+}}f(t)^{g(t)}$$
$$\lim_{t \rightarrow 0^{+}}g(t)\ln f(t)=\lim_{t \rightarrow 0^{+}}{(\ln \sin t})(\ln(2-\cos t))=\lim_{t \rightarrow 0^{+}}((1/2)t^2\ln t)=0=:L$$
Therefore:
$$\lim_{t \rightarrow 0^{+}}f(t)^{g(t)}=\lim_{t \rightarrow 0^{+}}\exp(L)=1$$
$\lim_{t \rightarrow 0^{-}} f(t)^{g(t)}$ can be considered in a similar fashion.
