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Compute the limit of the given function

Evaluate the limit.

    \[ \lim_{x \to \frac{1}{2}^-} \frac{\log(1-2x)}{\tan (\pi x)}. \]


Both the numerator and denominator go to 0 as x \to \frac{1}{2}^- so we apply L’Hopital’s rule,

    \begin{align*}  \lim_{x \to \frac{1}{2}^-} \frac{\log (1 - 2x)}{\tan (\pi x)} &= \lim_{x \to \frac{1}{2}^-} \frac{ \frac{-2}{1-2x} }{ \frac{\pi}{\cos^2 (\pi x)} } \\[9pt]  &= \lim_{x \to \frac{1}{2}^-} \frac{-2 \cos^2 (\pi x)}{\pi (1-2x)} \\[9pt]  &= \lim_{x \to \frac{1}{2}^-} \frac{(-4 \cos (\pi x)) (\pi) (- \sin (\pi x))}{-2 \pi} &(\text{L'Hopital's})\\[9pt]  &= 0. \end{align*}

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