Both the numerator and denominator go to 0 as so we apply L’Hopital’s rule,
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Eiji says:
The numerator does not go to 0, so how do you solve it?
S says:
It can be reduced to lim_(t -> 0+) sin(t) * log(t) / cos(t). The numerator there can be reduced to x * log(x), as x tends to 0. That can be transformed to -log(t)/t, as t tends to +infinity. Then apply T7.11 and you are done.
S says:
The initial limit can be transformed to $\lim_{x \to 0+} \frac{\sin x \log x}{\cos x}$. The denominator is 1, so let’s focus on the numerator. By application of Taylor’s expansion $\lim_{x \to 0+} (x + o(x)) \log x$. $\lim_{x \to 0+} x \log x = \lim_{t \to +\infty} \frac{-\log t}{t} = 0$ by T7.11. Similarly the limit for $o(x) \log x$ can be computed to be 0. Overall the numerator is 0, and the limit 0/1 = 0.
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![\[ \lim_{x \to \frac{1}{2}^-} \frac{\log(1-2x)}{\tan (\pi x)}. \]](https://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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-32c0db69c772d3ae48f68aa245f7795a_l3.png "Rendered by QuickLaTeX.com")
The numerator does not go to 0, so how do you solve it?
It can be reduced to lim_(t -> 0+) sin(t) * log(t) / cos(t). The numerator there can be reduced to x * log(x), as x tends to 0. That can be transformed to -log(t)/t, as t tends to +infinity. Then apply T7.11 and you are done.
The initial limit can be transformed to $\lim_{x \to 0+} \frac{\sin x \log x}{\cos x}$. The denominator is 1, so let’s focus on the numerator. By application of Taylor’s expansion $\lim_{x \to 0+} (x + o(x)) \log x$. $\lim_{x \to 0+} x \log x = \lim_{t \to +\infty} \frac{-\log t}{t} = 0$ by T7.11. Similarly the limit for $o(x) \log x$ can be computed to be 0. Overall the numerator is 0, and the limit 0/1 = 0.