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

Question

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

Answer 1

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

Answer 2

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 $b \ne 0$.

Answer 3

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:

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

Answer 4

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

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

Answer 5

(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))