Question Number 87690 by jagoll last updated on 05/Apr/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\mathrm{x}−\mathrm{sin}\:\mathrm{2x}}{\mathrm{x}−\mathrm{sin}\:\mathrm{x}} \\ $$
Commented by jagoll last updated on 05/Apr/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right)−\left(\mathrm{2x}−\frac{\mathrm{8x}^{\mathrm{3}} }{\mathrm{6}}\right)}{\mathrm{x}−\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{3}} }{\left(\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} \right)}\:=\:\mathrm{6} \\ $$
Commented by Ar Brandon last updated on 06/Apr/20
$${But}\:{when}\:{I}\:{replace}\:{x}\:{with}\:\mathrm{0}.\mathrm{00000001}\:{and}\:{perform} \\ $$$${the}\:{calculations}\:{it}\:{gives}\:\mathrm{0}.\:{Itsn}'{t}\:{that}\:{strange}? \\ $$
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
$${No}\:{big}\:{deal}.\:{The}\:{solution}\:{was}\:{to}\:{change}\:{the}\: \\ $$$${function}\:{to}\:{radians}.\:{Hum}! \\ $$
Answered by Ar Brandon last updated on 05/Apr/20
$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\frac{{d}^{\mathrm{3}} }{{dx}^{\mathrm{3}} }\left(\mathrm{2}{sin}\:{x}−{sin}\:\mathrm{2}{x}\right)}{\frac{{d}^{\mathrm{3}} }{{dx}^{\mathrm{3}} }\left({x}−{sin}\:{x}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{−\mathrm{2}{cos}\:{x}+\mathrm{8}{cos}\:\mathrm{2}{x}}{{cos}\:{x}}=\mathrm{8}−\mathrm{2}=\mathrm{6} \\ $$