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Question Number 87690 by jagoll last updated on 05/Apr/20

$$\underset{x\to 0}{\lim }\frac{2\sin \mathrm{x}-\sin 2\mathrm{x}}{\mathrm{x}-\sin \mathrm{x}}$$

Commented by jagoll last updated on 05/Apr/20

$$\underset{x\to 0}{\lim }\frac{2(\mathrm{x}-\frac{{\mathrm{x}}^3}{6})-(2\mathrm{x}-\frac{{8\mathrm{x}}^3}{6})}{\mathrm{x}-(\mathrm{x}-\frac{{\mathrm{x}}^3}{6})}=$$$$\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^3}{\left(\frac{1}{6}{\mathrm{x}}^3\right)}=6$$

Commented by Ar Brandon last updated on 06/Apr/20

$$ButwhenIreplacexwith0.00000001andperform$$$$thecalculationsitgives0.{Itsn}^{\prime }tthatstrange?$$

Commented by MJS last updated on 06/Apr/20

$$\mathrm{not}\mathrm{strange}\mathrm{at}\mathrm{all}.\mathrm{calculators}\mathrm{cannot}\mathrm{be}\mathrm{used}$$$$\mathrm{this}\mathrm{way},\mathrm{because}\mathrm{they}\mathrm{round}$$

Commented by Ar Brandon last updated on 08/Apr/20

$$Nobigdeal.Thesolutionwastochangethe$$$$functiontoradians.Hum!$$

Answered by Ar Brandon last updated on 05/Apr/20

$$=\underset{x\to 0}{lim}\frac{\frac{d^3}{{dx}^3}(2sinx-sin2x)}{\frac{d^3}{{dx}^3}(x-sinx)}$$$$=\underset{x\to 0}{lim}\frac{-2cosx+8cos2x}{cosx}=8-2=6$$

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