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Problems with calculating

$$\lim_{x\rightarrow0}\frac{\ln(\cos(2x))}{x\sin x}$$

$$\lim_{x\rightarrow0}\frac{\ln(\cos(2x))}{x\sin x}=\lim_{x\rightarrow0}\frac{\ln(2\cos^{2}(x)-1)}{(2\cos^{2}(x)-1)}\cdot \left(\frac{\sin x}{x}\right)^{-1}\cdot\frac{(2\cos^{2}(x)-1)}{x^{2}}=0$$

Correct answer is -2. Please show where this time I've error. Thanks in advance!

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You have $n\to 0$. I guess you want $x\to 0$. You might also want to use \left( and \right) for self adjusting parenthesis. – Pedro Tamaroff Jan 10 '13 at 18:01
You're using \lim wrong. – Asaf Karagila Jan 10 '13 at 18:03
The most obvious problem is that $\lim_{x\to 0}\big(2\cos^2x-1)=1$, so your last factor blows up. – Brian M. Scott Jan 10 '13 at 18:03
up vote 12 down vote accepted

The known limits you might wanna use are, for $x\to 0$

$$\frac{\log(1+x)}x\to 1$$

$$\frac{\sin x }x\to 1$$

With them, you get

$$\begin{align}\lim\limits_{x\to 0}\frac{\log(\cos 2x)}{x\sin x}&=\lim\limits_{x\to 0}\frac{\log(1-2\sin ^2 x)}{-2\sin ^2 x}\frac{-2\sin ^2 x}{x\sin x}\\&=-2\lim\limits_{x\to 0}\frac{\log(1-2\sin ^2 x)}{-2\sin ^2 x}\lim\limits_{x\to 0}\frac{\sin x}{x}\\&=-2\lim\limits_{u\to 0}\frac{\log(1+u)}{u}\lim\limits_{x\to 0}\frac{\sin x}{x}\\&=-2\cdot 1 \cdot 1 \\&=-2\end{align}$$

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$$\lim_{x\to\ 0}\frac{\log\cos 2x}{x\sin x}\stackrel{\text{L'Hospital}}=\lim_{x\to 0}\frac{-2\tan 2x}{\sin x+x\cos x}{}\stackrel{\text{L'H}}=\lim_{x\to 0}-\frac{4\sec 2x}{2\cos x-x\sin x}=-\frac{4}{2}=-2$$

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Nice Don, as always. +1 – Babak S. Jan 10 '13 at 19:53

I tink if you work as follows, you will get the answer better(I hope so):

When $\alpha(x)$ is very small, then $\ln(1+\alpha(x))\sim\alpha(x)$ so $$\ln\left(\cos(2x)\right)=\ln\left(1+(-2\sin^2(x)\right)\sim -2\sin^2(x)$$

Now take your limit with this fact again. It is $-2$.

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Nice your way (+1) – user 1618033 Feb 7 '13 at 11:13
@Chris'ssister: Thanks. it is kind of you. – Babak S. Feb 7 '13 at 11:36
Nicely done (+1)... – amWhy Feb 10 '13 at 14:40

$\lim_{x\to0}\cos 2x=1$ not $0$ unlike $\sin x$ so we can not write $\lim_{x\rightarrow0}\frac{\ln(2\cos^{2}(x)-1)}{(2\cos^{2}(x)-1)}$ to utilize $\lim_{y\to 0}\frac{\ln(1+y)}y=1$

Instead, we can try in the following way: $$\frac{\ln(1-2\sin^2x)}{x\sin x}=(-2)\frac{\ln(1-2\sin^2x)}{(-2\sin^2x)}\frac{\sin x}x$$

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As $x$ tends to $0$, $\cos(2x)$ tends to $1$. Hence, using $\ln(1+u) \sim u$, $\sin u \sim u$ and $1 - \cos u\sim \frac{u^2}{2}$, $$ \frac{\ln(\cos(2x))}{x\sin x} \sim \frac{\cos(2x)-1}{x\times x} \sim \frac{-\frac{(2x)^2}{2}}{x^2} \sim - 2 $$

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$$\lim_{x\to 0}\frac{\ln(\cos(2x))}{x\sin x}=\lim_{x\to 0}\frac{\ln(\cos(2x))}{1-\cos 2x}\times\lim_{x\to 0}\frac{x}{\sin x}\times\lim_{x\to 0}\frac{1-\cos 2x }{(2x)^2}\times4=-1\times1\times2=-2$$

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I think in your last fraction, it should be $x^2$ instead of $(2x)^2$ in the denominator, no? – Nils Matthes Feb 7 '13 at 11:09
@NilsMatthes: but then I multiplied all by $4$. – user 1618033 Feb 7 '13 at 11:12
oh, I see. Nevermind... – Nils Matthes Feb 7 '13 at 11:15

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