Can we prove that $\lim_{x \to 0} \frac{x - \sin x}{x^3} = \frac 16$ using just algebra, trigonometric theorems and notable limits $\left( \mathrm{i.e. }\quad \frac{\sin x}{x} \to 1 \quad \mathrm{and} \quad \frac{1 - \cos x}{x^2} \to \frac 12 \right) $ and no l'Hospital rule, no series expansion, no Taylor series?
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2https://math.stackexchange.com/questions/217081/determine-lim-x-to-0-fracx-sinxx3-frac16-without-lhospita?rq=1 – aleden Nov 22 '18 at 22:53
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More in general refer to Are all limits solvable without L'Hôpital Rule or Series Expansion – user Nov 22 '18 at 23:08
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$\sin x=\sin (x/3)\cos(2x/3)+\cos(x/3)\sin(2x/3)=\sin (x/3)(1-2\sin^2(2x/3))+2\cos^2(x/3)\sin(x/3)=\sin (x/3)-2\sin^3 (x/3)+2\sin (x/3)-2\sin^3 (x/3)=3\sin (x/3)-4\sin^3 (x/3)$
$\frac{x-\sin x}{x^3}=\frac{x-3\sin (x/3)-4\sin^3 (x/3)}{x^3}=\frac{x-3\sin (x/3)}{x^3}-\frac{4\sin^3 (x/3)}{x^3}=1/9\frac{x/3-\sin (x/3)}{(x/3)^3}-4/27 \frac{\sin^3 (x/3)}{(x/3)^3}$
$m=\lim_{x\rightarrow 0} \frac{x-\sin x}{x^3}= 1/9\lim_{x\rightarrow 0}\frac{x/3-\sin (x/3)}{(x/3)^3}-4/27\lim_{x\rightarrow 0} \frac{\sin^3 (x/3)}{(x/3)^3}=1/9M-4/27 \Rightarrow 8/9M=4/27\Rightarrow M=1/6.$
John_Wick
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This is a really nifty way to compute the limit, but it assumes the limit exists. – Barry Cipra Nov 22 '18 at 23:15
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