Friday, August 24, 2018

calculus - Solve $lim_{xto 0} frac{sin x-x}{x^3}$





I'm trying to solve this limit




$$\lim_{x\to 0} \frac{\sin x-x}{x^3}$$




Solving using L'hopital rule, we have:



$$\lim_{x\to 0} \frac{\sin x-x}{x^3}= \lim_{x\to 0} \frac{\cos x-1}{3x^2}=\lim_{x\to 0} \frac{-\sin x}{6x}=\lim_{x\to 0} \frac{-\cos x}{6}=-\frac{1}{6}.$$




Am I right?



I'm trying to solve this using change of variables, I need help.



Thanks



EDIT



I didn't understand the answer and the commentaries, I'm looking for an answer using change of variables.


Answer




I suppose the below counts as a change of variable.



Assuming that the limit exists, then you can compute the limit as follows:



Replace $x$ by $3x$, then the limit (say $L$) is



$$L = \lim_{x\to 0}\frac{\sin 3x - 3x}{27x^3} = \lim_{x\to 0}\frac{3\sin x - 3x - 4\sin^3 x}{27x^3} = $$
$$\lim_{x\to 0}\frac{1}{9}\left(\frac{\sin x - x}{x^3}\right) - \lim_{x\to 0}\frac{4}{27}\left(\frac{\sin^3 x}{x^3}\right)$$



(we used the formula $\sin 3x = 3\sin x - 4 \sin^3 x$).




Thus we get



$$L = \frac{L}{9} - \frac{4}{27} \implies L = -\frac{1}{6}$$



Of course, we still need to prove that the limit exists.


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