Evaluate a given limit Evaluate the following limit:
$$\lim_{x \rightarrow 4} \left( \frac{1}{13 - 3x} \right) ^ {\tan \frac{\pi x}{8}}$$
I haven't managed to get anything meaningful yet.
Thank you in advance!
 A: Set $x-4=h,13-3x=13-3(h+4)=1-3h$
$\tan\dfrac{\pi x}8=\tan\dfrac{\pi(4+h)}8=-\cot\dfrac{\pi h}8$
$$\dfrac1{1-3h}=1+\dfrac1{1-3h}-1=1+\dfrac{3h}{1-3h}$$
$$\lim_{h\to0}\left(1+\dfrac{3h}{1-3h}\right)^{-\cot(\pi h)/8}$$
$$=\left(\lim_{h\to0}\left(1+\dfrac{3h}{1-3h}\right)^{\dfrac{1-3h}{3h}}\right)^{-\lim_{h\to0}\dfrac{3h\cot(\pi h)/8}{1-3h}}$$
The inner limit converges to $e$
$$\lim_{h\to0}\dfrac{3h\cot(\pi h)/8}{1-3h}=3\lim_{h\to0}\dfrac{\cos(\pi h)/8}{1-3h}\cdot\dfrac1{\lim_{h\to0}\dfrac{\sin\pi h/8}{\pi h/8}}\cdot\dfrac8\pi=?$$
A: In this answer, the following is proven:
Lemma: Suppose $\lim\limits_{x\to0}xy(x)=a$, then
$$
\lim_{x\to0}(1+x)^y=e^a
$$
Therefore,
$$
\begin{align}
\lim_{x\to4}\left(\frac1{13-3x}\right)^{\large\tan\left(\frac{\pi x}8\right)}
&=\lim_{x\to0}\left(\frac1{1-3x}\right)^{\large\tan\left(\frac\pi2+\frac{\pi x}8\right)}\\
&=\lim_{x\to0}\left(1-3x\vphantom{\frac13}\right)^{\large1/\tan\left(\frac{\pi x}8\right)}\\[12pt]
&=e^{-24/\pi}
\end{align}
$$
Since
$$
\begin{align}
\lim_{x\to0}\frac{-3x}{\tan\left(\frac{\pi x}8\right)}
&=\lim_{x\to0}\frac{-3x}{\frac{\pi x}8}\lim_{x\to0}\frac{\frac{\pi x}8}{\tan\left(\frac{\pi x}8\right)}\\
&=-\frac{24}\pi\cdot1
\end{align}
$$
A: see my nice answer:
assume $x=t+4$ $$\lim_{x \rightarrow 4} \left( \frac{1}{13 - 3x} \right) ^ {\tan \frac{\pi x}{8}}=\lim_{t \to 0} \left( \frac{1}{13 - 3(t+4)} \right) ^ {\tan \frac{\pi (t+4)}{8}}$$
$$=\lim_{t \to 0} \left( \frac{1}{1-3t} \right) ^ {-\cot \frac{\pi t}{8}}$$
$$=\lim_{t \to 0} \left(1-3t \right) ^ {\cot \frac{\pi t}{8}}$$
above limit has form $1^{\infty}$, so
$$=e^{\large \lim_{t \to 0} \left(1-3t-1 \right) \cdot {\cot \frac{\pi t}{8}}}$$
$$=e^{\large \lim_{t \to 0} \left(-3t\right) \cdot {\cot \frac{\pi t}{8}}}$$
$$=e^{\large-3\cdot \frac{8}{\pi}\lim_{t \to 0}  \frac{\frac{\pi t}{8}}{\tan\frac{\pi t}{8}}}=e^{-24/\pi}$$
A: Compute the limit of the logarithm:
$$
\lim_{x\to4}\tan\frac{\pi x}{8}\log\frac{1}{13-3x}=
\lim_{x\to4}\frac{-\log(13-3x)}{\cot\frac{\pi x}{8}}
$$
This screams l’Hôpital!
$$
\lim_{x\to4}\frac{\frac{3}{13-3x}}{\frac{\pi}{8}(-1-\cot^2\frac{\pi x}{8})}=\frac{3}{-\pi/8}=-\frac{24}{\pi}
$$
Thus your limit is $e^{-24/\pi}$.
