Find the limit $\lim_{x\to 0} \frac{\ln(\cos (5x))}{\ln(\cos(4x))}$

Question

We have to use the $O(\cdot)$ notation while solving this problem. I have done several steps: $$ \lim_{x\to 0} \frac{\ln(\cos (5x))}{\ln(\cos(4x))} = \lim_{x\to 0} \frac{\ln \left(1 - \sin^2(5x/2)\right)} {\ln \left(1- \sin^2(2x)\right)}. $$ I don't know how to use the O notation after this

Answer 1

We have that

$$\frac{\ln(\cos (5x)}{\ln(\cos(4x)} =\frac{\ln(1+(\cos (5x)-1)}{\cos(5x)-1}\frac{\cos(4x)-1}{\ln(1+(\cos (4x)-1)}\frac{\cos(5x)-1}{(5x)^2}\frac{(4x)^2}{\cos(4x)-1} \frac{(5x)^2}{(4x)^2}$$

then refer to standard limits

• $\frac{\log(1+t)}t \to 1$
• $\frac{\cos t-1}{t^2} \to -\frac12$

or by Taylor's series we can use that

• $\cos(5x)=1-\frac{25}2x^2+O(x^4) \implies \ln(\cos(5x))=-\frac{25}2x^2+O(x^4)$
• $\cos(4x)=1-\frac{16}2x^2+O(x^4)\implies \ln(\cos(4x))=-\frac{16}2x^2+O(x^4)$

therefore

$$\frac{\ln(\cos (5x)}{\ln(\cos(4x)} =\frac{-\frac{25}2x^2+O(x^4)}{-\frac{16}2x^2+O(x^4)}=\frac{25+O(x^2)}{16+O(x^2)} \to \frac{25}{16}$$

Answer 2

Hint:

$$\lim_{x\to0}\dfrac{\ln(\cos2nx)}{x^2}=-2n^2\cdot\lim_{x\to0}\dfrac{\ln(1-2\sin^2nx)}{-2\sin^2nx}\left(\lim_{x\to0}\dfrac{\sin nx}{nx}\right)^2=-2n^2$$

Here $2n=5,4$

Alternatively $$\cos2nx=1-\dfrac{(2nx)^2}2+O(x^4)$$

Answer 3

One approach is to notice that as $x \to 0$, both numerator and denominator $\to \ln(1) = 0$ as well, so L'Hospital's Rule applies.

Let $f_a(x) = \ln \cos (ax)$ then $$ f_a'(x) = \frac{d[\ln \cos (ax)]}{dx} = \frac{\frac{d}{dx} [\cos ax]}{\cos ax} = \frac{-a\sin (ax)}{\cos(ax)} $$

So you need to compute $$ \lim_{x \to 0} \frac{f_a(x)}{f_b(x)}, $$ can you finish this now?

Answer 4

Use the O($\cdot$) notation for $\sin t = t + O(t^3)$ and $\ln (1-u)= -u + O(u^2)$ to express,

$$\ln (1-\sin^2 t) = -\sin^2 t +O(\sin^4 t ) = -[t + O(t^3)]^2 + O(t^4) = -t^2 + O(t^4)$$

Then, continue the limit you have,

$$\lim_{x\to 0} \frac{\ln \left(1 - \sin^2\frac{5x}{2}\right)}{\ln \left(1- \sin^2 2x\right)} =\lim_{x\to 0} \frac{-\left(\frac{5x}2\right)^2 + O(x^4)}{-(2x)^2+ O(x^4) }=\frac{\frac{25}{4}x^2}{4x^2}=\frac{25}{16}$$