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

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.

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    $\begingroup$ Could you use the big O? $\endgroup$ – Ma Ming Apr 29 '14 at 22:43
  • $\begingroup$ Perhaps this will help you: $\cos(3x)=4\cos^3(x)-3\cos(x)$ $\endgroup$ – barak manos Apr 29 '14 at 23:23
  • $\begingroup$ @barakmanos, excellent idea -- see the "added later" part of my answer. $\endgroup$ – Barry Cipra Apr 30 '14 at 1:47
  • $\begingroup$ No idea what the big O is :D $\endgroup$ – Kinan Apr 30 '14 at 18:09
  • $\begingroup$ Possible duplicate of Limit evaluate $\lim_{x\to0}{{\frac{\ln(\cos(4x))}{\ln(\cos(3x))}}}$? $\endgroup$ – Arnaud D. Dec 5 '19 at 11:01
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$\quad$If you are allowed to use the fact that $\ln(1+t)\approx t$ when $t\to0$, then write:

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

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

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    $\begingroup$ To conclude that $\lim_{\theta\to0}{\sin\theta\over\theta}=1$ is to assume that the area of a circle is circle $\pi r^2$, and vice versa. Another alternative is to use L'hosital's Rule, but to find the derivative of $\sin x$ we need to evaluate the limit. Does anyone else get the feeling that we are just going in circles. $\endgroup$ – John Joy Apr 30 '14 at 0:27
  • $\begingroup$ @JohnJoy: The context of this and many other similar questions on this site is the following: Apparently in some (American?) schools, people are taught the value of certain “remarkable” limits before they are taught the tools by which they are deduced. Kinda like being told that “to be or not to be” is a famous line from the famous play “Hamlet” written by the famous dramaturge Shakespeare long before actually studying either in literature class. (Hey, it makes no sense to me either, but that's life). :-) $\endgroup$ – Lucian Apr 30 '14 at 7:53
  • $\begingroup$ I'm allowed to use that limit of sinx/x but I don't know that limit of ln(1+t) so I'm not allowed to use it :( so I need another way without using ln(1+t) limit. Anyway, in my textbook they gave a hint about adding 1 and subtracting 1 in each ln I tried to do so and got something like ln(cos^2(x/2)-1)/ln(cos^2(3x/2)-1) and couldn't go on with it cause I didn't know what to do next $\endgroup$ – Kinan Apr 30 '14 at 18:16
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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).

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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 $\displaystyle{\lim_{x\rightarrow0}{\frac{\ln(\cos x)}{\ln(\cos 3x)}}=\frac{1}{9}}$.

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  • $\begingroup$ Well, I haven't got a clue about what you did, I'm not that advanced in calculus, I've been studying it for couple years only but thanks anyway because that'll make sense to me few years later:) $\endgroup$ – Kinan Apr 30 '14 at 18:08

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