Find the limit $ \lim_{x \to (\frac{1}{2})^{-}} \frac{\ln(1 - 2x)}{\tan \pi x} $ Question
$$
  \lim_{x \to (\frac{1}{2})^{-}} \frac{\ln(1 - 2x)}{\tan \pi x} 
$$
I'm not sure how to go about this limit, I've tried to apply L'Hopital's rule
(as shown).
It seems that the form is going to be forever indeterminate? Unless I'm missing
something, but the limit is (apparently) zero.
Working
As this is in the indeterminate form of $- \infty / + \infty$, apply L'Hopital's rule.
Let $\frac{\ln(1 - 2x)}{\tan \pi x} = f/g$, then
$f' = \frac{-2}{1 - 2x}$ , and  $f'' = \frac{-4}{(1 - 2x)^2}$ 
$g' = \pi \sec^2 \pi x$ (which is equal to $\pi (\tan^2 (\pi x )+ 1)$ ) and
$g'' = 2 \pi^2 \sec^2 (\pi x) \tan \pi x $ ( or
$2 \pi^2 (\tan^2 (\pi x) + 1) \tan \pi x$
)
Using the first derivatives gives
$$
  \lim_{x \to (\frac{1}{2})^{-}}
  \frac{\frac{-2}{1 - 2x}}{\pi \sec^2 \pi x}
$$
Which is in the form
$$
  \frac{- \infty}{+ \infty}
$$
So that I would now use the second derivative, which is 
$$
  \lim_{x \to (\frac{1}{2})^{-}}
  \frac{
    \frac{-4}{(1 - 2x)^2}
  }{
    2 \pi^2 \sec^2 (\pi x) \tan \pi x
  }
$$
Or 
$$
  \lim_{x \to (\frac{1}{2})^{-}}
  \frac{
    \frac{-4}{(1 - 2x)^2}
  }{
    2 \pi^2 (\tan^2 (\pi x) + 1) \tan \pi x
  }
$$
But this is still in an indeterminate form?
 A: Let $x-\frac{1}{2}=y$. Hence, $x=\frac{1}{2}+y$ and
 $$\lim_{x \to \frac{1}{2}^{-}} \frac{\ln(1 - 2x)}{\tan \pi x} =-\lim_{y\rightarrow0^-}\frac{\ln(-y)}{\cot{\pi y}}=\pi\lim_{y\rightarrow0^-}\left(-y\ln(-y)\right)=0$$
A: As $x<\dfrac12\iff2x-1<0$
set $1-2x=y$
$$F=\lim_{y\to0^+}\dfrac{\ln y}{\tan\dfrac{\pi(1-y)}2}=\lim_{y\to0^+}\dfrac{\ln y}{\cot\dfrac{\pi y}2}$$
Applying L'Hospital's $$F=\lim_{y\to0^+}\dfrac{\sin^2\dfrac{\pi y}2}{-\dfrac\pi2\cdot y}\cdots=0$$  using $\lim_{h\to0}\dfrac{\sin h}h=1$
A: You were doing fine:

Using the first derivatives gives
  $$  \lim_{x \to (\frac{1}{2})^{-}}
  \frac{\frac{-2}{1 - 2x}}{\pi \sec^2 \pi x}$$

Now simplify a bit first:
$$\frac{\frac{-2}{1 - 2x}}{\pi \sec^2 \pi x} = \frac{2\cos^2\pi x}{\pi \left(2x-1\right)}$$
Applying l'Hôpital to the fraction in this form gives the easy $-\sin\left( 2\pi x\right)$, which evaluates to $0$.
A: You do not need de l'Hospital rule for the evaluation of such limit:
$$\lim_{x\to(1/2)^{-}}\frac{\log(1-2x)}{\tan(\pi x)}=\lim_{z\to 0^+}\frac{\log(2z)}{\cot(\pi z)}=\lim_{z\to 0^+}\frac{z\log(2z)}{z\cot(\pi z)} =\frac{0}{\frac{1}{\pi}}=0.$$
It is enough to exploit a substitution $x\mapsto \frac{1}{2}-z$ and the well-known limits (that should be studied before approaching de l'Hospital rule, the Stolz-Cesàro theorem and so on)
$$ \lim_{x\to 0}\frac{\sin x}{x}=1,\qquad \forall\epsilon>0,\;\;\lim_{x\to 0^+} x^{\varepsilon}\log(x)=0.$$
A: Let $y = \displaystyle \frac{1}{2} - x$. As $x \to \displaystyle \Big(\frac{1}{2}\Big)^{-}$, $y \to 0^{+}$. Then:
$$ \lim \limits_{x \to \Big( \frac{1}{2} \Big)^{-}} \frac{\ln(1-2x)}{\tan(\pi x)} = \lim \limits_{x \to 0^{+}} \frac{\ln(2y)}{\tan\Big( \frac{\pi}{2} - \pi y \Big)} $$
Because $\displaystyle \tan\Big( \frac{\pi}{2} - \pi y \Big) = \frac{1}{\tan(\pi y)}$, we have:
$$ \lim \limits_{x \to 0^{+}} \frac{\ln(2y)}{\tan\Big( \frac{\pi}{2} - \pi y \Big)} = \lim \limits_{x \to 0^{+}} \ln(2y) \tan(\pi  y). $$
As $y \to 0$, $\tan(\pi y) \sim \pi y$. Therefore:
$$ \ln(2y) \tan(\pi y) \mathop{\sim} \limits_{y \to 0^{+}} \pi y \ln(2y). $$
As a consequence :

$$ \lim \limits_{x \to \Big( \frac{1}{2} \Big)^{-}} \frac{\ln(1-2x)}{\tan(\pi x)} = 0. $$

