What do you think about my solution on this limit: $\lim_{x\to 0}\frac{2x-\sin(2x)}{x^3}=\frac{4}{3}$ We make the substitution $x=3y$ $$\lim_{x\rightarrow 0}\frac{2x-\sin 2x}{x^{3}}=\lim_{y\rightarrow 0}\frac{6y-\sin 6y}{27y^{3}}=\lim_{y\rightarrow 0}\frac{6y-3\sin 2y}{27y^{3}}+\lim_{y\rightarrow 0}\frac{4(\sin 2y)^{3}}{27{y^{3}}}$$ We get $$l=\frac{3}{27}l+\frac{32}{27}$$ and $$l=\frac{4}{3}$$
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1You did not prove that there was a limit. It is fine apart from that. – Aphelli Jan 09 '19 at 14:43
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How can you write $\sin 6y=3\sin 2y$? – snowfall512 Jan 09 '19 at 14:44
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@pie314271 He's using the triple angle formula for $\sin$ – saulspatz Jan 09 '19 at 14:45
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(Oh, I thought that the plus sign at the end of the first line was another equals sign for some reason. Sorry.) – snowfall512 Jan 09 '19 at 14:47
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https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion – lab bhattacharjee Jan 09 '19 at 15:06
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This looks fine, except that you didn't prove your original limit exists. But this is a rather complicated way to derive this limit, you could have simply used L'Hospital's rule twice, or a simple Taylor expansion in the form $\sin(2x) = 2x - \frac{(2x)^3}{6} + o(x^3)$.
A. Bailleul
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