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$$\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin3x\cdot\ln(1+\sin4x)}$$

My steps

$\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin3x\cdot\ln(1+\sin4x)} = \lim_{x\rightarrow0}\frac{1-\cos2x}{2x}\cdot2x\cdot\frac{3x}{\sin3x}\cdot\frac{1}{3x}\cdot\frac{1}{\ln(1+\sin4x)}\cdot\frac{4x\sin4x}{4x\sin4x}\\= \lim_{x\rightarrow0}\frac{2x}{3x\cdot4x}\cdot\frac{1-\cos2x}{2x}\cdot\frac{3x}{\sin3x}\cdot\frac{4x}{\sin4x}\cdot\frac{\sin4x}{\ln(1+\sin4x)} = \lim_{x\rightarrow0}\frac{2}{12x}$

I'm so close to the right answer $\frac{1}{6}$.

What did I do wrong?

Edit 1: So apparently, $\lim_{x\rightarrow0}\frac{1-\cos2x}{2x} = 0 \not=1$ and so the limit would be $\lim_{x\rightarrow0}\frac{2x}{3x\cdot4x}\cdot\frac{1-\cos2x}{2x}\cdot\frac{3x}{\sin3x}\cdot\frac{4x}{\sin4x}\cdot\frac{\sin4x}{\ln(1+\sin4x)} = 0\cdot\frac{2}{12\cdot0}$

And so now the limit is $[\frac{0}{0}]$. I'm not sure how to fix this.

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  • $\begingroup$ I managed to get $\frac{1}{6}$ by using L'Hospital twice then substituting $x=0$. The terms get quite big and messy but it works. $\endgroup$ – Jam Nov 4 '14 at 15:32
  • $\begingroup$ Yeah, but I haven't learnt L'Hopitals rule yet :/ $\endgroup$ – B. Lee Nov 4 '14 at 15:32
  • $\begingroup$ I don't really understand your reasoning though. I'll agree that you can cancel the $\frac{3x}{\sin(3x)}$ terms but how have you taken out the term with $\frac{2x}{3x\cdot4x}$ & the term with the log? $\endgroup$ – Jam Nov 4 '14 at 15:36
  • $\begingroup$ $\lim_{x\rightarrow0}\frac{\ln(1+x)}{x} = 1$. And the thing is, $\frac{2x}{3x\cdot4x}$ is my problem right now. $\endgroup$ – B. Lee Nov 4 '14 at 15:37
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    $\begingroup$ Actually, you could do it by rearranging the term with $\cos$ and the term with $\frac{2x}{3x\cdot4x}$ to $2\cdot\frac{1-\cos(2x)}{(2x)^2}$ which can be shown to be $2\cdot\frac{1}{2}$ without L'Hospital (see this link enotes.com/homework-help/calculate-limit-1-cosx-x-2-231157 ). Then the limits should all be $1$ & you'll be left with $\frac{1}{6}$ $\endgroup$ – Jam Nov 4 '14 at 15:46
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We have $1-\cos(2x)\sim \frac{(2x)^2}{2}=2x^2$ as $x\to 0$; $\sin(3x)\sim 3x$, $\ln(1+\sin(4x))\sim \sin(4x)\sim 4x$ as $x\to 0$. Thus $$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {2x} \right)}}{{\sin \left( {3x} \right).\ln \left( {1 + \sin \left( {4x} \right)} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{x^2}}}{{3x.4x}} = \frac{1}{6}.$$

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  • $\begingroup$ I'm not sure I understand how $\lim_{x\rightarrow0}(1-\cos2x) = \frac{(2x)^2}{2}$, or the others for that matter. I appreciate the work, but I'm trying to use the common limits: en.wikipedia.org/wiki/List_of_limits $\endgroup$ – B. Lee Nov 4 '14 at 15:41
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    $\begingroup$ No. $1-cos(2x)\sim \frac{(2x)^2}{2}$ as $x\to 0$ means that $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {2x} \right)}}{{\frac{{{{\left( {2x} \right)}^2}}}{2}}} = 1$. In fact $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {2x} \right)}} {{\frac{{{{\left( {2x} \right)}^2}}} {2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x}} {{2{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}} {{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}} {x}.\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}} {x} = 1.$ $\endgroup$ – User3101 Nov 4 '14 at 15:52
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    $\begingroup$ Definition: $f(x)\sim g(x)$ as $x\to a$ if $\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = 1$. $\endgroup$ – User3101 Nov 4 '14 at 15:54

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