# Find the limit $\lim\limits_{x\to 0} \frac{\ln (\cos ax)}{\ln (\cos bx)}$ [closed]

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

-

## closed as off-topic by Thursday, PVAL, Tunk-Fey, Antonio Vargas, Claude LeiboviciAug 15 at 5:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Thursday, PVAL, Tunk-Fey, Antonio Vargas, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

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 Jul 15 '12 at 10:19

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

-
@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 $b \ne 0$.

-

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:

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

-
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 $\lim_{x\to0} \frac{{1-\cos x}}{x^2}=\frac{1}{2}$:

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

-

(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))

-
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