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now I am evaluating limits of functions, but i dont know how to start to solve this limit. It is possible without L Hopital's rule?

$\lim_{x\to0}{{\frac{\ln(\cos(4x))}{\ln(\cos(3x))}}}$?

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  • $\begingroup$ Any good reason not to use L'Hôpital's rule? This limit is crying out for it. $\endgroup$
    – Simon S
    Commented Nov 4, 2015 at 18:27
  • $\begingroup$ Our professor has forbidden this rule. $\endgroup$
    – Patrik Dendis
    Commented Nov 4, 2015 at 18:32
  • $\begingroup$ Are you allowed to use a power series expansion of $\ln(1 + x)$? $\endgroup$
    – Simon S
    Commented Nov 4, 2015 at 18:33
  • $\begingroup$ See also this question (asked one day earlier): math.stackexchange.com/questions/1511813/… $\endgroup$
    – Hans Lundmark
    Commented Jan 8, 2020 at 14:08

5 Answers 5

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By using Chebyshev polynomials of the first kind,

$$ \lim_{x\to 0}\frac{\log(\cos(4x))}{\log(\cos(3x))}=\lim_{t\to 1}\frac{\log(8t^3-8t^2+1)}{\log(4t^3-3t)}=\lim_{z\to 0}\frac{\log(1+16 z+40 z^2+32 z^3+8 z^4)}{\log(1 + 9 z + 12 z^2 + 4 z^3)}$$ hence the wanted limit equals $\large\color{red}{\frac{16}{9}}$.

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  • $\begingroup$ No offense @Jack, but don't you think that chebyshevs are a litte bit heavy for this kind of question ;) $\endgroup$
    – tired
    Commented Nov 4, 2015 at 18:39
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    $\begingroup$ @tired: no offense taken, you are clearly right, but if we have our hands tied, we have to be creative :D $\endgroup$
    – Jack D'Aurizio
    Commented Nov 4, 2015 at 18:40
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One option (if you can use power series, which require at least as much calculus as L'Hopital's rule!):

In any sufficiently small neighborhood of $ x = 0 $, $\cos (ax) = \sqrt{1 - \sin^2(ax)}$. Thus the original quotient equals

$$\frac{\ln(1 - \sin^2(4x))}{\ln(1 - \sin^2(3x))} = \frac{ -\sin^2(4x) + O(x^4)}{-\sin^2(3x) + O(x^4)} = \frac{ -\sin^2(4x)/x^2 + O(x^2)}{-\sin^2(3x)/x^2 + O(x^2)} \to \frac{4^2}{3^2}$$

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  • $\begingroup$ Can you help what is 'O' ? $\endgroup$
    – Patrik Dendis
    Commented Nov 4, 2015 at 18:56
  • $\begingroup$ $O(x^n)$ means the rest of the terms are $x^n$ or above and hence $O(x^4)/x^2 = O(x^2) \to 0$ as $x \to 0$. $\endgroup$
    – Simon S
    Commented Nov 4, 2015 at 18:59
  • $\begingroup$ If this is notation you don't have, you need to help us understand what tools you can use. $\endgroup$
    – Simon S
    Commented Nov 4, 2015 at 18:59
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Without Taylor or L'Hopital: We know $\lim_{u\to 0} [\ln (1+u)]/u = 1.$ (This is just $\ln'(1)=1.$) Thus

$$\ln (\cos u) = \frac{\ln (1+(\cos u-1))}{\cos u -1}(\cos u -1)$$ $$ =\frac{\ln (1+(\cos u-1))}{\cos u -1}\frac{1}{\cos u +1}(\cos^2 u -1)$$ $$ = \frac{\ln (1+(\cos u-1))}{\cos u -1}\frac{1}{\cos u +1}(-\sin^2 u)$$

As $u \to 0,$ this looks like $1\cdot (1/2)\cdot (-\sin^2 u).$ Use this on both $\ln (\cos 4x), \ln (\cos 3x).$

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With equivalents:

$\ln(u)\sim_1 u-1$, hence $\enspace\dfrac{\ln\cos4x}{\ln\cos3x}\sim_0\dfrac{\cos4x-1}{\cos3x-1}\sim_0\dfrac{-\dfrac{16x^2}2}{-\dfrac{9x^2}2}=\dfrac{16}9. $

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I'm going to combine the proposed methods.

$$f(x)=\ln(x)\implies\mathcal D_f=(0,+\infty)\tag 1$$

$$\begin{aligned}\ln(\cos\alpha) &=\ln\left(\sqrt{1-\sin^2\alpha }\right)\\&=\frac12\ln\left(1-\sin^2\alpha\right),\\&\ \ \ \alpha\in\left(-\frac{\pi}2+2k\pi,\frac{\pi}2+2k\pi\right),\ k\in\Bbb Z\end{aligned}\tag 2$$ $$\begin{aligned}&\color{white}{=}\lim_{x\to0}\frac{\ln(\cos(4x))}{\ln(\cos(3x))}\\&=\lim_{x\to0}\frac{\frac12\ln\left(1-\sin^2(4x)\right)}{\frac12\ln\left(1-\sin^2(3x)\right)}\\&=\lim_{x\to0}\frac{-\frac{\ln\left(1-\sin^2(4x)\right)}{\sin^2(4x)}\frac{\sin^2(4x)}{16x^2}16x^2}{-\frac{\ln\left(1-\sin^2(3x)\right)}{\sin^2(3x)}\frac{\sin^2(3x)}{9x^2}9x^2}\\&=\frac{16}9\end{aligned}$$

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