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$$\lim_{x \to 0} \frac{1-\cos(\sin(4x))}{\sin^2(\sin(3x))}$$

How can I evaluate this limit without using the L'Hospital Rule? I've expanded $\sin(4x)$ as $\sin(2x+2x)$, $\sin(3x) = \sin(2x + x)$, but none of these things worked.

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The simplest way is to note $\sin x \simeq x$ for small $x$ and $\cos x \simeq 1-\frac{x^2}{2}$ for small $x$. Then, you can obtain $$\frac{1-\cos\sin 4x}{\sin^2\sin 3x} \simeq \frac{1-\cos 4x}{\sin^2 3x} \simeq \frac{8x^2}{9x^2} = \frac{8}{9}.$$ You need to use Taylor series to formalize this type of argument.

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  • $\begingroup$ The cosine one is probably the main one you really want Taylor series for, by the way. Many calculus textbooks these days provide some sort of geometric proof that $\lim\limits_{x\rightarrow 0}\sin(x)/x = 1$, which translates to $\sin(x)\simeq x$ for small $x$, and $\lim\limits_{x\rightarrow 0} \frac{1-\cos(x)}{x}=0$. Most professors don't cover these proofs in much or any detail, because they are long, but they are often in the book. The issue with the cosine one is that it doesn't tell you about the 1/2 part. $\endgroup$ – zibadawa timmy Nov 6 '13 at 1:52
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You can start by writing

$${1-\cos(\sin(4x))\over\sin^2(\sin(3x))}= {1-\cos(\sin(4x))\over\sin^2(4x)} \left({\sin(4x)\over4x}\right)^2 \left({4x\over3x}\right)^2 \left({3x\over\sin(3x)}\right)^2 \left({\sin(3x)\over\sin(\sin(3x))}\right)^2$$

Now note that

$$\lim_{x\rightarrow0}{1-\cos(\sin(4x))\over\sin^2(4x)}=\lim_{u\rightarrow0}{1-\cos u\over u^2}={1\over2}$$

$$\lim_{x\rightarrow0}{\sin(4x)\over4x}=\lim_{u\rightarrow0}{\sin u\over u}=1$$

and so forth for the others. This leads to

$$\lim_{x\rightarrow0}{1-\cos(\sin(4x))\over\sin^2(\sin(3x))}={1\over2}\left(1\right)^2\left({4\over3}\right)^2(1)^2(1)^2={8\over9}$$

At the very least you can write

$$\lim_{x\rightarrow0}{1-\cos(\sin(4x))\over\sin^2(\sin(3x))}={16\over9}\lim_{u\rightarrow0}{1-\cos u\over u^2}\lim_{u\rightarrow0}{u\over\sin u}$$

(assuming the two limits on the right hand side exist).

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  • $\begingroup$ I like this one. I thought there was a trigonometric solution, but couldn't see it. $\endgroup$ – Betty Mock Nov 7 '13 at 1:32
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Start with a Taylor's polynomial. $cos x = 1 -x^2/2! + o(x^4)$. $sinx = x -x^3/3! +o(x^6)$. The o's just mean that the remaining terms are like $x^4$ and $x^6$. For purposes of seeing what happens at x = 0 they are irrelevant. With this we have $sin^2x = x^2 +o(x^4).$

So $(1-cosx)/sin^2(x) \approx (x^2/2)/x^2 = 1/2. \hspace{50px}$(1)

Since you were asked about sin3x and sin4x let's put those into (1).

$(1-cos(sin4x)/sin^2(sin3x) \approx [(sin^2(4x))/2]/sin^23x$.

By the computations above $sin^24x \approx (4x)^2 + o(x^4)$ and $sin^23x \approx (3x)^2 + o(x^4)$ . Putting it together we have

$[(sin^2(4x))/2]/sin^23x \approx [(4x)^2/2]/(3x)^2 =[16x^2/2]/9x^2 = 8/9$

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  • 1
    $\begingroup$ This is exactly Lord Soth's answer $\endgroup$ – Alex Nov 6 '13 at 1:35
  • $\begingroup$ He posted his answer while I was typing, so I didn't know it was there. Lord Soth does say that one has to work it out with a Taylor's series, and you see the work out in my answer. $\endgroup$ – Betty Mock Nov 7 '13 at 1:30

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