# Is it possible to solve this limit without Hopital / Taylor / derivatives?

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)?

• See the beautiful answer by moderator robjohn 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 4:10

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

• @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
• 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 Oct 24 '15 at 19:07
• 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 Oct 25 '15 at 5:47