Решить предел lim при x->0 от (ln(cos3x) / ln(cos5x)). Created by tarasgorodchukowd1fp. algebra-ru

ln(cos5x))...

Munkush $1)\; \; \; \; (ln(cos3x))'=\frac{1}{cos3x}\cdot (-sin3x)\cdot 3=-3tg3x\\\\(ln(cos5x))'=-5tg5x\\\\2)\; \; \; (-3tg3x)'=-3\cdot \frac{1}{cos^23x}\cdot 3=-\frac{9}{cos^23x}\\\\(-5tg5x)'=-\frac{25}{cos^25x}\\\\3)\; \; \; \lim\limits _{x \to 0} \frac{ln(cos3x)}{ln(cos5x)}= \lim\limits _{x \to 0}\frac{-3tg3x}{-5tg5x}= \lim\limits _{x \to 0} \frac{-\frac{9}{cos^23x}}{-\frac{25}{cos^25x}}=[\; cos0=1\; ]=\\\\= \frac{9}{25}=0,36$

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$\lim _{x\to \:0}\left(\frac{\ln \left(\cos \left(3x\right)\right)}{\ln \left(\cos \left(5x\right)\right)}\right)$

Применим правило Лопиталя
$\lim _{x\to \:0}\left(\frac{-3\tan \left(3x\right)}{-\frac{5\sin \left(5x\right)}{\cos \left(5x\right)}}\right)=\lim _{x\to \:0}\left(\frac{3\tan \left(3x\right)\cos \left(5x\right)}{5\sin \left(5x\right)}\right) \\ \\ \\ =\lim _{x\to \:0}\left(\frac{3\left(3\sec ^2\left(3x\right)\cos \left(5x\right)-5\sin \left(5x\right)\tan \left(3x\right)\right)}{25\cos \left(5x\right)}\right) \\ \\ \\$

$=\lim _{x\to \:0}\left(\frac{3\left(3\cos \left(5x\right)-5\cos ^2\left(3x\right)\sin \left(5x\right)\tan \left(3x\right)\right)}{25\cos ^2\left(3x\right)\cos \left(5x\right)}\right)= \\ \\ \\ =\frac{3\left(3\cos \left(5\cdot \:0\right)-5\cos ^2\left(3\cdot \:0\right)\sin \left(5\cdot \:0\right)\tan \left(3\cdot \:0\right)\right)}{25\cos ^2\left(3\cdot \:0\right)\cos \left(5\cdot \:0\right)}=\frac{9}{25}=0,36$

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