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$$\lim_{x \to 0}\frac{\cos(x) - 8x\sin(x/2) - \cos(3x)}{x^4}$$ I think I should replace by equivalent, such as $\sin(x)$ ~ $x$, but got nothing.

Thank you for answers, but what about solving without using L'Hôpital's rule?

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  • $\begingroup$ Do you know L'Hôpital's rule? $\endgroup$ – rajb245 Oct 29 '15 at 16:05
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    $\begingroup$ You need more terms in Taylor expansions. $\endgroup$ – Claude Leibovici Oct 29 '15 at 16:05
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    $\begingroup$ The limit appears to be $-19/6$. Numerically $-19/6=-3.1666...$ and at $x=.01$ the expression is $-3.16657..$. $\endgroup$ – coffeemath Oct 29 '15 at 16:12
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$\cos(x)-\cos(3x) = 2\sin(x)\sin(2x) = 4\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\sin(2x)$, hence we want to compute:

$$ \lim_{x\to 0}\frac{4\sin\left(\frac{x}{2}\right)}{x}\cdot\frac{\sin(2x)-2x}{x^3}=2\cdot\lim_{x\to 0}\frac{\sin(2x)-2x}{x^3}=16\cdot\lim_{z\to 0}\frac{\sin(z)-z}{z^3} $$ that equals $-\frac{8}{3}$, by applying twice De l'Hopital theorem, then subtract:

$$ \lim_{x\to 0}\frac{4\sin\left(\frac{x}{2}\right)}{x}\cdot\frac{\sin(2x)(1-\cos\frac{x}{2})}{x^3}=4\cdot\lim_{x\to 0}\frac{\sin(2x)\sin^2\left(\frac{x}{4}\right)}{x^3}=\frac{1}{2}. $$ The given limit is so $-\frac{8}{3}-\frac{1}{2}=\color{red}{\Large -\frac{19}{6}}$.


To prove the crucial part, i.e. $\lim_{x\to 0}\frac{x-\sin(x)}{x^3}=\frac{1}{6}$, without derivatives, you may assume that the limit just exists and equals $L$. Then: $$ L = \lim_{x\to 0}\frac{2x-\sin(2x)}{8x^3} = \lim_{x\to 0}\frac{x-\sin(x)\cos(x)}{4x^3}=\frac{L}{4}+\lim_{x\to 0}\frac{\sin(x)\sin^2\left(\frac{x}{2}\right)}{2x^3} $$ and that leads to $L=\frac{L}{4}+\frac{1}{8}$, from which $L=\frac{1}{6}$.

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  • $\begingroup$ Thank you, but I need to solve it without derivatives and L'Hopital. Could you help me? $\endgroup$ – AnatoliySultanov Oct 29 '15 at 16:35
  • $\begingroup$ @AnatoliySultanov: the problem boils down to proving that $$\lim_{x\to 0}\frac{x-\sin x}{x^3}=\frac{1}{6}.$$ What are the allowed tools? $\endgroup$ – Jack D'Aurizio Oct 29 '15 at 16:39
  • $\begingroup$ I can use O-notation and asymptotical equivalence of functions, but I can't use Taylor series and L'Hopital. $\endgroup$ – AnatoliySultanov Oct 29 '15 at 16:45
  • $\begingroup$ @AnatoliySultanov: I added an approach through a change a variable that just requires to show that the limit $\lim_{x\to 0}\frac{x-\sin x}{x^3}$ exists. Anyway, to use $\sin(x)=x-\frac{x^3}{6}+O(x^5)$ is not much different from using De l'Hopital theorem. How do you prove $\sin(x)=x-\frac{x^3}{6}+O(x^5)$ otherwise? $\endgroup$ – Jack D'Aurizio Oct 29 '15 at 16:48
  • $\begingroup$ +1 for the manipulation argument that gets the crucial limit assuming it exists. I don't think I've seen this before, but perhaps it appears in one of the references at the bottom of this 22 April 2008 post in the ap-calculus discussion group, archived at Math Forum. $\endgroup$ – Dave L. Renfro Oct 29 '15 at 16:59
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You need to consider all three terms of the numerator at the same time, as they are all needed to cancel.

You know that the numerator has a power series expansion, so let $n$ be the numerator. Then, either by direct computation, or by expanding each term, we have $n(0) = n'(0) = n''(0) = n'''(0) = 0$, $n''''(0) = - {19 \over 6}$.

Hence $n(x) = - {19 \over 6}x^4 + x^5 r(x)$, where $r$ is bounded near zero, hence we see the limit is $- {19 \over 6}$.

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