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Question Number 163825 by cortano1 last updated on 11/Jan/22

lim_(x→0) ((tan 2x−2x)/(sin 3x−3x)) =?

$$\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{2}{x}−\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{3}{x}−\mathrm{3}{x}}\:=?\: \\ $$

Answered by Ar Brandon last updated on 11/Jan/22

lim_(x→0) ((tan2x−2x)/(sin3x−3x))=lim_(x→0) (((2x+(((2x)^3 )/3)−2x))/(3x−(((3x)^3 )/(3!))−3x)) =−lim_(x→0) (((8x^3 )/3)×(6/(27x^3 )))=−((16)/(27))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan2}{x}−\mathrm{2}{x}}{\mathrm{sin3}{x}−\mathrm{3}{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2}{x}+\frac{\left(\mathrm{2}{x}\right)^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2}{x}\right)}{\mathrm{3}{x}−\frac{\left(\mathrm{3}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}−\mathrm{3}{x}} \\ $$$$=−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}}×\frac{\mathrm{6}}{\mathrm{27}{x}^{\mathrm{3}} }\right)=−\frac{\mathrm{16}}{\mathrm{27}} \\ $$

Answered by bobhans last updated on 11/Jan/22

= (8/(27)) lim_(x→0) ((tan 2x−2x)/((2x)^3 )) . lim_(x→0) (((3x)^3 )/(sin 3x−3x)) = (8/(27)) ×(1/3)×(−6)=−((16)/(27))

$$\:=\:\frac{\mathrm{8}}{\mathrm{27}}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{2x}−\mathrm{2x}}{\left(\mathrm{2x}\right)^{\mathrm{3}} }\:.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{3x}\right)^{\mathrm{3}} }{\mathrm{sin}\:\mathrm{3x}−\mathrm{3x}}\: \\ $$$$\:=\:\frac{\mathrm{8}}{\mathrm{27}}\:×\frac{\mathrm{1}}{\mathrm{3}}×\left(−\mathrm{6}\right)=−\frac{\mathrm{16}}{\mathrm{27}} \\ $$

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