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How can I calculate the following limit without using, as Wolfram Alpha does, without using l'Hôpital? $$ \lim_{x\to 0}\frac{\log\cos x}{\log\cos 3x} $$

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It would be nice if you would format this. –  ncmathsadist Jun 27 '12 at 0:00
    
yes.....thanks. –  Alan Turing Jun 27 '12 at 0:07
    
thank you "Dylan Moreland" –  Alan Turing Jun 27 '12 at 0:13

4 Answers 4

up vote 2 down vote accepted

We need some limiting property of $\log(x)$. The only limiting property we will use is that $$ \lim_{u\to0}\frac{\log(1+u)}{u}=1 $$ With $1+u=\cos^2(x)$, we get $$ \begin{align} \lim_{x\to0}\frac{\log(\cos(x))}{\log(\cos(3x))} &=\lim_{x\to0}\frac{\log(\cos(x))}{\log(4\cos^3(x)-3\cos(x))}\\ &=\lim_{x\to0}\frac{\log(\cos(x))}{\log(4\cos^2(x)-3)+\log(\cos(x))}\\ &=\lim_{x\to0}\frac{\frac12\log(\cos^2(x))}{\log(4\cos^2(x)-3)+\frac12\log(\cos^2(x))}\\ &=\lim_{u\to0}\frac{\log(1+u)}{2\log(1+4u)+\log(1+u)}\\[4pt] &=\frac{\lim\limits_{u\to0}\frac{\log(1+u)}{u}}{8\lim\limits_{u\to0}\frac{\log(1+4u)}{4u}+\lim\limits_{u\to0}\frac{\log(1+u)}{u}}\\[4pt] &=\frac{1}{8+1}\\[4pt] &=\frac19 \end{align} $$

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\begin{align} \dfrac{\log(\cos(x))}{\log(\cos(3x))} & = \dfrac{2\log(\cos(x))}{2\log(\cos(3x))} \\ &= \dfrac{\log(\cos^2(x))}{\log(\cos^2(3x))}\\ & =\dfrac{\log(1-\sin^2(x))}{\log(1-\sin^2(3x))}\\ & =\dfrac{\log(1-\sin^2(x))}{\sin^2(x)} \times \dfrac{\sin^2(3x)}{\log(1-\sin^2(3x))} \times \dfrac{\sin^2(x)}{\sin^2(3x)}\\ & =\dfrac{\log(1-\sin^2(x))}{\sin^2(x)} \times \dfrac{\sin^2(3x)}{\log(1-\sin^2(3x))} \times \dfrac{\sin^2(x)}{x^2} \times \dfrac{(3x)^2}{\sin^2(3x)} \times \dfrac19 \end{align} Now recall the following limits $$\lim_{y \to 0} \dfrac{\log(1-y)}{y} = -1$$ $$\lim_{z \to 0} \dfrac{\sin(z)}{z} = 1$$ Also, note that as $x \to 0$, $\sin(kx) \to 0$. Hence, $$\lim_{x \to 0} \dfrac{\log(1-\sin^2(x))}{\sin^2(x)} = -1$$ $$\lim_{x \to 0} \dfrac{\sin^2(3x)}{\log(1-\sin^2(3x))} = -1$$ $$\lim_{x \to 0} \dfrac{\sin^2(x)}{x^2} = 1$$ $$\lim_{x \to 0} \dfrac{(3x)^2}{\sin^2(3x)} = 1$$ Hence, $$\lim_{x \to 0} \dfrac{\log(\cos(x))}{\log(\cos(3x))} = (-1) \times (-1) \times 1 \times 1 \times \dfrac19 = \dfrac19$$ Hence, the limit is $\dfrac19$.

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oh ...my god thank you Marvis –  Alan Turing Jun 27 '12 at 1:31

Another way is via the approximations $\cos(x) = 1-\frac{x^2}{2}+O(x^4)$ and $\log(1+x) = x+O(x^2)$:

$$\begin{align}\frac{\log(\cos(x))}{\log(\cos(3x))} &= \frac{\log(1-\frac{x^2}{2}+O(x^4))}{\log(1-\frac{(3x)^2}{2}+O(x^4))} \\ &= \frac{-\frac{x^2}{2}+O(x^4)}{-\frac{(3x)^2}{2}+O(x^4)} \\ &= \frac{-\frac{x^2}{2}}{-\frac{x^2}{2}}\frac{1+O(x^2)}{9+O(x^2)} \\ &= \frac{1}{9}+O(x^2)\end{align}$$

And now the limit as $x\rightarrow0$ is trivial.

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Let's see a fast & elementary way:

$$\lim_{x\to 0}\frac{\log\cos x}{\log\cos 3x}=\lim_{x\to 0}\frac{\cos x -1}{\cos 3x - 1} = \lim_{x\to 0}\frac{\frac{\cos x -1}{x^2}}{\frac{\cos 3x - 1}{9x^2}}\cdot\frac{x^2}{9x^2}=\frac{1}{9}.$$

Q.E.D.

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@Alan Turing: i used the fact that $\lim_{x\to0} \frac{\ln(1+x)}{x}=1$ and $\lim_{x\to0} \frac{1-\cos x}{x^2}=\frac{1}{2}$. –  Chris's sis Jun 27 '12 at 5:43
    
thank you Chris –  Alan Turing Jun 27 '12 at 18:33

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