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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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    thank you "Dylan Moreland"2012-06-27

5 Answers 5

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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 $\lim_{x\to0} \frac{1-\cos x}{x^2}=\frac{1}{2}$.2012-06-27
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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 Marvis2012-06-27
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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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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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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 $\begin{aligned}L &= \lim_{x \to 0}\frac{\log \cos x}{\log \cos 3x}\\ &= \lim_{x \to 0}\frac{\log \cos x}{\cos x - 1}\cdot\frac{\cos x - 1}{\cos 3x - 1}\cdot\frac{\cos 3x - 1}{\log \cos 3x}\\ &= \lim_{x \to 0}\frac{\cos x - 1}{\cos 3x - 1}\\ &= \lim_{x \to 0}\frac{(\cos x - 1)(\cos x + 1)}{\cos x + 1}\cdot\frac{\cos 3x + 1}{(\cos 3x - 1)(\cos 3x + 1)}\\ &= \lim_{x \to 0}\frac{1 + \cos 3x}{1 + \cos x}\cdot\frac{\sin^{2}x}{\sin^{2}3x}\\ &= \lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot \frac{x^{2}}{(3x)^{2}}\cdot\frac{(3x)^{2}}{\sin^{2}3x}\\ &= \frac{1}{9}\end{aligned}$