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It's simple to prove with Hopital that

$$ \lim_{x \to 0} \frac{x-\sin(x)}{x^3} = \frac{1}{6}$$

Is it possible to solve this limit without Hopital or Taylor (without derivatives)?

Martin Sleziak
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arulbero
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    See the beautiful answer by moderator robjohn http://math.stackexchange.com/a/438121/72031 From the answer it should be very obvious that the problem is very difficult if one fordbids the use of differentiation. This somehow shows the great power of differentiation. – Paramanand Singh Oct 26 '15 at 04:10
  • https://math.stackexchange.com/a/158134/1242 – Hans Lundmark Jan 08 '20 at 14:00

1 Answers1

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Let us assume $$\displaystyle \lim_{x\rightarrow 0}\frac{x-\sin x}{x^3} = L$$ (A finite quantity).

Now replace $x\rightarrow 3y$, then we get $$\displaystyle \lim_{y\rightarrow 0}\frac{3y-\sin 3y}{27y^3} = L$$

Now, using the formula $$\sin 3y = 3\sin y-4\sin^3 y$$

we get $$\displaystyle \lim_{y\rightarrow 0}\frac{3y-3\sin y+4\sin^3 y}{27y^3} = L$$

So $$\displaystyle \frac{1}{9}\lim_{y\rightarrow 0}\frac{y-\sin y}{y^3}+\frac{4}{27}\displaystyle \lim_{y\rightarrow 0}\left(\frac{\sin y}{y}\right)^3=L$$

So $$\frac{1}{9}L+\frac{4}{27} = L\Rightarrow \frac{8}{9}L = \frac{4}{27}\Rightarrow L=\frac{1}{6}$$

psmears
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juantheron
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  • @juantheron This is a very interesting approach indeed. +1) Did you replace $x=3y$ because there is a cube in the denominator? I am trying to come up with some generalization... – imranfat Oct 24 '15 at 18:37
  • Yes imranfat You are Right. – juantheron Oct 24 '15 at 18:39
  • @juantheron Thank you very much! – arulbero Oct 24 '15 at 18:45
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    What is really proved in this great solution is that if we assume that the limit exists then it equals to 1/6. The existence of the limits remains unproved. – Idris Addou Oct 24 '15 at 19:07
  • Is there a way to show "that it is bounded near $0$" "without Hopital or Taylor (without derivatives)"? $\hspace{.44 in}$ –  Oct 24 '15 at 21:41
  • @RickyDemer You can probably use the same techniques (namely trig identities) that you use to show that $\lim_{x\rightarrow 0} \sin(x)/x = 1$, but I've not tried to work that out to see if it is correct. – zibadawa timmy Oct 24 '15 at 22:14
  • I found another answer link – arulbero Oct 24 '15 at 22:34
  • First show that $$\lim_{x\to0} \frac{x-sin(x)}{x-tan(x)}=-\frac{1}{2}$$ then subtract 1 $$\lim_{x\to0} \frac{tan(x)-sin(x)}{tan(x)-x}=\frac{3}{2}$$ then $$\lim_{x\to0} \frac{tan(x)-x}{x^3}=\frac{1}{3}$$ and $$\lim_{x\to0} \frac{x-sin(x)}{x^3}=\frac{1}{6}$$ – arulbero Oct 24 '15 at 22:50
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    If one consider any function defined arround zero then exactly one situation is true among thé following four situations. 1) the limit is +infinity 2) the limit is - infinity 3) the limit is some réal number 4) the limit is none of the above (so it do not existe as a réal number nor as +- infinity) the proof above is that if the possibility number 3) holds then that real number mentionned in that possibility is 1/6. It remains to prove that possibilités 1), 2) and 4) do not occur for the considéred function. – Idris Addou Oct 25 '15 at 05:47