How can I calculate $\lim_{x \to 0} \log(\cos(x))/\log(\cos(3x))$ without l'Hopital?

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

-
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

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}

-

\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 $\dfrac19$.

-
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.

-

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.

-
@Alan Turing: i used the fact that $\lim_{x\to0} \frac{\ln(1+x)}{x}=1$ and $\lim_{x\to0} \frac{1-\cos x}{x^2}=\frac{1}{2}$. – user 1618033 Jun 27 '12 at 5:43
thank you Chris – Alan Turing Jun 27 '12 at 18:33

Note that since $x \to 0$ both $\cos x$ and $\cos 3x$ tend to $1$ and then using $\lim\limits_{t \to 1}\dfrac{\log t}{t - 1} = 1$ we get $$\lim_{x \to 0}\frac{\log\cos x}{\cos x - 1} = 1 = \lim_{x \to 0}\frac{\log \cos 3x}{\cos 3x - 1}$$ We can now easily see that \begin{aligned}L &= \lim_{x \to 0}\frac{\log \cos x}{\log \cos 3x}\\ &= \lim_{x \to 0}\frac{\log \cos x}{\cos x - 1}\cdot\frac{\cos x - 1}{\cos 3x - 1}\cdot\frac{\cos 3x - 1}{\log \cos 3x}\\ &= \lim_{x \to 0}\frac{\cos x - 1}{\cos 3x - 1}\\ &= \lim_{x \to 0}\frac{(\cos x - 1)(\cos x + 1)}{\cos x + 1}\cdot\frac{\cos 3x + 1}{(\cos 3x - 1)(\cos 3x + 1)}\\ &= \lim_{x \to 0}\frac{1 + \cos 3x}{1 + \cos x}\cdot\frac{\sin^{2}x}{\sin^{2}3x}\\ &= \lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot \frac{x^{2}}{(3x)^{2}}\cdot\frac{(3x)^{2}}{\sin^{2}3x}\\ &= \frac{1}{9}\end{aligned}

-