Solving limit without using L'Hopital

Question

$$\lim_{x\to 0} \frac{\ln(\cos x)}{\ln(\cos 3x)}$$

The answer is $1/9$, but I don't want to use L'Hopital because I'm not supposed to.

Any help would be greatly appreciated and thanks in advance.

Answer 1

$\quad$If you are allowed to use the fact that $\ln(1+t)\approx t$ when $t\to0$, then write:

$$\lim_{x\to0}\frac{\ln\cos x}{\ln\cos3x}=\lim_{x\to0}\frac{\ln\sqrt{1-\sin^2x}}{\ln\sqrt{1-\sin^23x}}=\lim_{x\to0}\frac{\frac12\cdot\ln(1-\sin^2x)}{\frac12\cdot\ln(1-\sin^23x)}=\lim_{x\to0}\frac{\ln(1-\sin^2x)}{\ln(1-\sin^23x)}=$$

$$=\lim_{x\to0}\frac{-\sin^2x}{-\sin^23x}=\lim_{x\to0}\frac{\sin^2x}{\sin^23x}=\lim_{x\to0}\frac{\sin^2x}{(3x)^2}\cdot\frac{(3x)^2}{\sin^23x}=\frac19\quad,\quad\text{since}\quad\lim_{x\to0}\frac{\sin x}x=1.$$

Answer 2

A little bit depends on what you are allowed to use. Note that

$${\ln\cos x\over\ln\cos3x}={\displaystyle{\ln\cos x\over1-\cos x}\over\displaystyle{\ln\cos3x\over1-\cos3x}}\cdot{\sin^2x\over\sin^23x}\cdot{1+\cos3x\over1+\cos x}$$

So if we are allowed to use

$$\lim_{\theta\to0}{\sin\theta\over\theta}=1$$

we easily have

$$\lim_{x\to0}{\sin^2x\over\sin^23x}={1\over 9}\lim_{x\to0}{\left({\sin x\over x}\right)^2\over\left({\sin 3x\over 3x}\right)^2}={1\over9}$$

and if we are allowed to use

$$\lim_{u\to1}{\ln u\over1-u}=-1$$

then we can wrap things up as

$$\lim_{x\to0}{\ln\cos x\over\ln\cos3x}={\displaystyle\lim_{x\to0}{\ln\cos x\over1-\cos x}\over\displaystyle\lim_{x\to0}{\ln\cos3x\over1-\cos3x}}\cdot\lim_{x\to0}{\sin^2x\over\sin^23x}\cdot\lim_{x\to0}{1+\cos3x\over1+\cos x}={-1\over-1}\cdot{1\over9}\cdot{2\over2}={1\over9}$$

Actually we don't need to know the value of the limits for $\sin\theta/\theta$ and $\ln u/(1-u)$, just that those limits exist and are nonzero.

Added later: The suggestion made by barak manos in comments to use the identity $\cos3x=4\cos^3x-3\cos x$ is an excellent one. It can be used to eliminate the need to know anything about the limit of $\sin x/x$.

It's convenient to invert the fraction to

$$\lim_{x\to0}{\ln\cos3x\over\ln\cos x}=\lim_{x\to0}{\ln\cos x+\ln(4\cos^2x-3)\over\ln\cos x}=1+\lim_{c\to1}{\ln(4c^2-3)\over\ln c}$$

If we now write

$${\ln(4c^2-3)\over\ln c}=4(1+c)\left({\displaystyle{\ln(4c^2-3)\over1-(4c^2-3)}\over\displaystyle{\ln c\over1-c}}\right)$$

we can conclude that

$$\lim_{x\to0}{\ln\cos3x\over\ln\cos x}=1+8{\displaystyle\lim_{u\to1}{\ln u\over1-u}\over\displaystyle\lim_{u\to1}{\ln u\over1-u}}=9$$

(and thus the original limit is $1/9$), provided we know that $\lim_{u\to1}{\ln u\over1-u}$ exists and is nonzero.

It's kind of nice that you can get rid of all the trig before worrying about trying to compute the limit. I don't see any way, offhand, to get rid of the logarithm (without replacing it with an equally hard to handle exponential function).

Answer 3

Since $\ln(x)=\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}(x-1)^n}{n!}}}$ and $\cos x=\displaystyle{\sum_{n=0}^{\infty}{\frac{(-1)^nx^{2n}}{(2n)!}}}=1-\frac{x^2}{2}+\frac{x^4}{24}+\ldots$ we can estimate $\ln(\cos x)$ and $\ln(\cos 3x)$ as follows \begin{align} \ln(\cos x)&=-\frac{1}{2}x^2+O(x^4) &\text{and} & &\ln(\cos 3x)&=-\frac{1}{2}(3x)^2+O(x^4)=-\frac{9}{2}x^2+O(x^4) \\ \end{align} where $O(x^4)$ means that all remaining terms are of order $4$ and greater. So \begin{align} \frac{\ln(\cos x)}{\ln(\cos 3x)}&=\frac{-\frac{1}{2}x^2+O(x^4)}{-\frac{9}{2}x^2+O(x^4)} \\ &= \frac{-\frac{1}{2}x^2[1+O(x^2)]}{-\frac{1}{2}x^2[9+O(x^2)]}\\ &= \frac{1+O(x^2)}{9+O(x^2)}\\ \end{align} Hence $\displaystyle{\lim_{x\rightarrow0}{\frac{\ln(\cos x)}{\ln(\cos 3x)}}=\frac{1}{9}}$.