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$$\lim_{x \to 0} \frac{9x(\cos4x-1)}{\sin8x-8x}$$

I have done this problem a couple of times and could not get the correct answer. Here is the work I have done so far http://imgur.com/GDZjX26 . The correct answer was $\frac{27}{32}$, did I differentiate wrong somewhere?

Kot
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    $$\sin(8x)=8x-\frac{256x^3}{6}+O(x^5)$$ $$\cos(4x)=1-8x^2+O(x^4)$$ For sufficiently small $x$ – Ethan Splaver Apr 28 '13 at 04:55
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    In your lines 2 & 3, when you differentiated $-4x \sin 4x$, you produced a term $+16x \cos 4x$. You didn't want to "flip the sign" when you differentiated sine... – colormegone Apr 28 '13 at 04:57
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    Typically, calculus students learn L'Hôpital's rule in calc I, and Taylor polynomials in calc II. Sadly, indeterminate forms are often not taught as an application of the latter. – Ted Shifrin Apr 28 '13 at 04:58
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    @Ted Shifrin: Very much agree. L'Hospital's Rule has a mechanical character, while I think of the Maclaurin polynomial as confronting the behaviour of the function near $0$. – André Nicolas Apr 28 '13 at 05:05

2 Answers2

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You have to use L'Hopitals 3 times we have $$\begin{align} \lim_{x\to 0}\frac{9x(\text{cos}(4x)-1}{\text{sin}(8x)-8x}&=\lim_{x\to 0}\frac{(9 (\text{cos}(4 x)-1)-36 x \text{sin}(4 x))}{(8 \text{cos}(8 x)-8)}\\ &=\lim_{x\to 0}\frac{-1}{64}\frac{(-72 \text{sin}(4 x)-144 x \text{cos}(4 x))}{\text{sin}(8x)}\\&=\lim_{x\to 0}\frac{-1}{512}\frac{(576 x \text{sin}(4 x)-432 \text{cos}(4 x))}{\text{cos}(8x)}\\&=\frac{432}{512}\\&=\frac{27}{32}.\end{align}$$

i.a.m
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Using L'Hospital Rule repeatedly,

$$\lim_{x \to 0} \frac{9x(\cos4x-1)}{\sin8x-8x}$$

$$=\lim_{x \to 0} \frac{9(\cos4x-1)+9x(-4\sin4x)}{8\cos8x-8}$$

$$=\frac98\left( \lim_{x \to 0} \frac{\cos4x-1}{\cos8x-1}\right)-\frac92\left( \lim_{x \to 0}\frac{x\sin4x}{\cos8x-1} \right)$$

$$(1)\lim_{x \to 0} \frac{\cos4x-1}{\cos8x-1}=\lim_{x \to 0} \frac{-4\sin4x}{-8\sin8x}=\frac12\lim_{x \to 0} \frac{\sin4x}{\sin8x}=\frac12\lim_{x \to 0} \frac{4\cos4x}{8\cos8x}=\frac12\frac48$$

$$(2)\lim_{x \to 0}\frac{x\sin4x}{\cos8x-1}= \lim_{x \to 0}\frac{\sin4x+4x\cos4x}{-8\sin8x}$$ $$=-\frac18\lim_{x \to 0}\frac{\sin4x}{\sin8x}-\frac12 \lim_{x \to 0}\frac{x}{\sin8x}\cdot \lim_{x \to 0}\frac{\cos4x}1$$

$$=-\frac18\cdot\frac48 \text{(already found)}-\frac12 \lim_{x \to 0}\frac1{8\cos8x}\cdot1$$

$$=-\frac1{16}-\frac12\cdot\frac18=-\frac18 $$

Can you take it home form here?

Alternatively,

using $\cos2y=1-2\sin^2y,$

$$(1)\lim_{x \to 0} \frac{\cos4x-1}{\cos8x-1}=\left(\lim_{x \to 0} \frac{\sin2x}{\sin4x}\right)^2$$

Now, $$\lim_{x \to 0} \frac{\sin2x}{\sin4x}=\lim_{x \to 0} \frac{2\cos2x}{4\cos4x}=\frac24=\frac12$$

$$(2) \lim_{x \to 0}\frac{x\sin4x}{\cos8x-1}=\lim_{x \to 0}\frac{x\sin4x}{-2\sin^24x}$$ $$=-\lim_{x \to 0}\frac{x}{2\sin4x}\text{ as } x\to0,\sin4x\to0\implies \sin4x\ne0$$ $$=-\lim_{x \to 0}\frac1{2\cdot4\cos4x}=-\frac18$$

lab bhattacharjee
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