Home » Blog » Compute the limit of the given function Compute the limit of the given function by RoRi December 25, 2015 Evaluate the limit. Both the numerator and denominator go to 0 as so we apply L’Hopital’s rule, Related
![\[ \lim_{x \to \frac{1}{2}^-} \frac{\log(1-2x)}{\tan (\pi x)}. \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e51cbf24c64eb54dc1403c3e8e458b8c_l3.png "Rendered by QuickLaTeX.com")
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*}](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-32c0db69c772d3ae48f68aa245f7795a_l3.png "Rendered by QuickLaTeX.com")
so we apply L’Hopital’s rule,
The numerator does not go to 0, so how do you solve it?