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$\lim_{x\to0}\frac{\sin x - x}{x^3}$

I know it can be easily done by using Taylor expansion of sine function and L'Hopital. However, can we come up with a way to solve the limits using properties.

Plus Twenty
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Mike Karter
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  • See also here: https://math.stackexchange.com/questions/157903/evaluation-of-lim-limits-x-rightarrow0-frac-tanx-xx3/158134#158134 – Hans Lundmark Jun 12 '19 at 11:12

1 Answers1

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Let $x = 3y$

$\sin(x) = \sin(3y) = 3\sin(y) - 4\sin^3(y)$

$$L = \lim_{y\to0}\frac{3\sin(y) - 4\sin^3(y) - 3y}{(3y)^3} = \lim_{y\to0}\frac{\sin(y) + y}{9y^3}+\lim_{y\to0}\frac{- 4\sin^3(y) }{27y^3}$$

$$L = \frac{L}{9} - \frac{4}{27} $$

$$\frac{8L}{9} = - \frac{4}{27} $$

$$L = \frac{-1}{6}$$

19aksh
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    Well done, but it actually doesn't really prove the existence of the limit. But in the case where you proved that it exist, then it's well done :) – user657324 Jun 12 '19 at 08:24