5
$\begingroup$

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)?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Paramanand Singh Oct 26 '15 at 4:10
18
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @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... $\endgroup$ – imranfat Oct 24 '15 at 18:37
  • $\begingroup$ Yes imranfat You are Right. $\endgroup$ – juantheron Oct 24 '15 at 18:39
  • $\begingroup$ @juantheron Thank you very much! $\endgroup$ – arulbero Oct 24 '15 at 18:45
  • 12
    $\begingroup$ 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. $\endgroup$ – Idris Oct 24 '15 at 19:07
  • 1
    $\begingroup$ 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. $\endgroup$ – Idris Oct 25 '15 at 5:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.