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I need to solve this limit without using L'Hopital's rule. I have attempted to solve this countless times, and yet, I seem to always end up with an equation that's far more complicated than the one I've started with.

$$ \lim_{x\to0} \frac{\ln(\cos5x)}{\ln(\cos7x)}$$

Could anyone explain me how to solve this without using L'Hopital's rule ? Thanks!

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Hint use the facts that $$\frac{\ln x}{x-1}\to 1$$ and $$\frac{1-\cos x }{x^2}\to \frac{1}{2}$$

Merry Christmas.

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  • $\begingroup$ Thanks ^_^ , Merry Christmas to you as well :)) $\endgroup$ – Stefan Dec 25 '16 at 21:36
  • $\begingroup$ But how do you get that without L'Hopital? $\endgroup$ – skan Dec 25 '16 at 22:41
  • $\begingroup$ @skan Can't you work it out ? I don't feel like writing more today, but look at some of my old answers to similar questions, you'll see the method. $\endgroup$ – Rene Schipperus Dec 25 '16 at 23:18
  • $\begingroup$ @skan The first one is just $\ln'(1) = 1.$ The second one follows from $\sin x/x \to 1;$ undoubtedly you have seen this before. $\endgroup$ – zhw. Dec 25 '16 at 23:21
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    $\begingroup$ @skan Sorry. Do you mean how do you get the limits I quoted ? They are basic, you need these limits in order to prove the derivatives of log and sin are what they are, so using hospital on them is circular. $\endgroup$ – Rene Schipperus Dec 25 '16 at 23:33
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HINT:

Note that we have

$$\cos(n x)=1-2\sin^2(nx/2)$$

and

$$ \log(\cos(nx))=\frac{\log(1-2\sin^2(nx/2)}{2\sin^2(nx/2)}\, \frac{2\sin^2(nx/2)}{(nx/2)^2} (nx/2)^2$$

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$$x \to 0 \to \ln(1+x)\sim x\\ x \to 0 \to cos(ax)\sim1-\frac{1}{2}(ax)^2\\ \lim_{x \to 0}\frac{\ln (\cos 5x)}{\ln (\cos 7x)}=\\ \lim_{x \to 0}\frac{\ln (1-\frac{1}{2}(5x)^2)}{\ln (1-\frac{1}{2}(7x)^2)}=\\ \lim_{x \to 0}\frac{-\frac{1}{2}(5x)^2}{-\frac{1}{2}(7x)^2}=\\\frac{25}{49}$$

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  • $\begingroup$ $u~v$ doesn't implie $\ ln(u)~\ln(v).$ $\endgroup$ – hamam_Abdallah Dec 25 '16 at 21:07
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We have $$\ln(\cos(X))=\ln(1+(\cos(X)-1))\sim \cos(X)-1\sim \frac{-X^2}{2}\;(X\to 0)$$

thus the limit is $$\frac{25}{49}.$$

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You can use Cauchy's mean value theorem, but in fact that's kind of cheating, as this is how L' Hopital's rule is proved.

For every $x$ around 0 there exists by Cauchy's mean value theorem a $\vert \xi_x \vert \leq \vert x \vert$ such that

$$\frac{\ln(\cos(5x))}{\ln(\cos(7x))} = \frac{\ln(\cos(5x)) - \ln(\cos(0))}{\ln(\cos(7x))- \ln(\cos(0))} = \frac{\left( \frac{-5\sin(5\xi_x)}{\cos(5\xi_x)} \right)}{\left( \frac{-7\sin(7\xi_x)}{\cos(7\xi_x)} \right)} = \frac{5}{7}\frac{\cos(7\xi_x)}{\cos(5\xi_x)}\cdot \frac{\sin(5\xi_x)}{\sin(7\xi_x)}.$$

As $\xi_x\rightarrow 0$ for $x\rightarrow 0$, we get (if the respective limits exist)

$$ \lim_{x\rightarrow 0} \frac{\ln(\cos(5x))}{\ln(\cos(7x))} = \lim_{\xi \rightarrow 0} \frac{5}{7}\frac{\cos(7\xi_x)}{\cos(5\xi_x)}\cdot \frac{\sin(5\xi_x)}{\sin(7\xi_x)} =\lim_{\xi \rightarrow 0} \frac{5}{7} \frac{\sin(5\xi_x)}{\sin(7\xi_x)}.$$

Repeating the argument for $\frac{\sin(5\xi_x)}{\sin(7\xi_x)}$ yields that the limit equals $\frac{25}{49}$.

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