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Help me calculate this limit: $\lim\limits_{x\to 0} \frac{\ln (\cos ax)}{\ln (\cos bx)}$

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closed as off-topic by 1999, PVAL, Tunk-Fey, Antonio Vargas, Claude Leibovici Aug 15 '14 at 5:39

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What have you tried? –  Jonas Meyer Jul 15 '12 at 8:07
L'Hopital's rule, along with the chain rule perhaps? –  user22805 Jul 15 '12 at 8:09
Well, it doesn't seem to me a bad problem at all. In fact, it's a nice one for beginners ... (+1) –  Chris's sis the artist Jul 15 '12 at 10:19

5 Answers 5

$$ \lim\limits_{x\to 0} \frac{\ln (\cos ax)}{\ln (\cos bx)}= \lim\limits_{x\to 0}\frac{(1-\cos ax)\ln \Big((1+\cos ax -1)^\frac{1}{\cos ax -1}\Big)}{(1-\cos bx)\ln\Big((1+\cos bx -1)^\frac{1}{\cos bx -1}\Big)} =\lim_{x\to 0}\frac{\frac{1-\cos ax}{(ax)^2}}{\frac{1-\cos bx}{(bx)^2}}\cdot\left(\frac{ax}{bx}\right)^2=\frac{a^2}{b^2} $$

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@Cocopuffs: The derivation is right, too. $\lim_{x\to0}\ln(1+\cos ax-1)^{\frac1{\cos ax-1}}=\ln e=1$, so you have $\lim_{x\to0}\frac{1-\cos ax}{1-\cos bx}$, which you multiply and divide by $\left(\frac{ax}{bx}\right)^2$. Then you use the limit $\lim_{x\to0}\frac{1-\cos t}{t^2}=\frac12$. A little explanation would have helped, but it’s actually rather clever. +1 –  Brian M. Scott Jul 15 '12 at 8:55
@BrianM.Scott Aha! I thought the $\ln$ was inside the exponent. Removing comment. –  Cocopuffs Jul 15 '12 at 9:01
@Cocopuffs: I’ve taken the liberty of adding some parentheses to make it clearer. –  Brian M. Scott Jul 15 '12 at 9:21

Use L'Hospital's rule twice to get

$$\lim_{x \rightarrow 0} \frac{\log(\cos ax)}{\log(\cos bx)} = \lim_{x \rightarrow 0} \frac{a \tan(ax)}{b \tan(bx)} = \lim_{x \rightarrow 0} \frac{a^2 \sec^2(ax)}{b^2 \sec^2(bx)} = \frac{a^2}{b^2}$$ for $b \ne 0$.

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You need to use the chain rule twice. The answer is $a^2/b^2$. Differentiating twice and taking the limit as x goes to zero gives:

$$\lim _{x\rightarrow 0} \left( -{a}^{2}-{\frac { \left( \sin \left( ax \right) \right) ^{2}{a}^{2}}{ \left( \cos \left( ax \right) \right) ^{2}}} \right) \left( -{b}^{2}-{\frac { \left( \sin \left( b x \right) \right) ^{2}{b}^{2}}{ \left( \cos \left( bx \right) \right) ^{2}}} \right) ^{-1}$$

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You need to enclose LaTeX formulas in dollar signs $-$ single dollar signs for in-line formulas, double for displayed formulas. –  Brian M. Scott Jul 15 '12 at 8:44

Let's use some elementary limits, namely $\lim_{x\to0} \frac{\ln{(1+x)}}{x}=1$ and $\lim_{x\to0} \frac{{1-\cos x}}{x^2}=\frac{1}{2}$:

$$\lim\limits_{x\to 0} \frac{\ln [1+(\cos ax-1)]}{\ln [1+(\cos bx-1)]}\cdot \frac{\cos bx-1}{\cos ax-1}\cdot \frac{\cos ax-1}{\cos bx-1}=\lim\limits_{x\to 0} \frac{\cos ax-1}{\cos bx-1} \cdot \frac{{(bx)}^2}{{(ax)}^2}\cdot \frac{{(ax)}^2}{{(bx)}^2} = \frac{-\frac{1}{2}}{{-\frac{1}{2}}}\cdot\frac{{a}^2}{{b}^2}$$ $$ L = \frac{{a}^2}{{b}^2}.$$

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(ln(x))'= x'/x so.. lim x=0 ln(cos(ax))=-asin(ax)/cos(ax)= -asin(ax)*cos(bx)= a^2 =a /ln(cos(bx)) /-bsin(bx)/cos(bx) /-bsin(bx)*cos(ax) /b^2 /b

request to be edited by an operator so it can be easilly viewed ps it's two lines, the second starts from /ln(cos(bx))

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I cannot edit it, as I don't understand what you want to say and why it's a useful addition over the other answers. –  martini Nov 19 '12 at 14:46
I used the rule that says that (ln(fx))'= fx'/fx. I think it's simpler than the rest. I just cannot properly complete it nor edit it, so I ask from a more experienced one –  Andreas Nov 19 '12 at 18:48
when in edit it somekind shows the right form –  Andreas Nov 19 '12 at 19:27

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