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Question Number 126389 by shaker last updated on 20/Dec/20

Answered by liberty last updated on 20/Dec/20

 lim_(x→0)  (( [((3cos 3x)/(sin 3x))])/( [((2cos 2x)/(sin 2x))])) = lim_(x→0) ((3cos 3x)/(sin 3x))×((sin 2x)/(2cos 2x))  = (3/2)×lim_(x→0)  ((sin 2x)/(sin 3x)) = 1

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\:\left[\frac{\mathrm{3cos}\:\mathrm{3}{x}}{\mathrm{sin}\:\mathrm{3}{x}}\right]}{\:\left[\frac{\mathrm{2cos}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{2}{x}}\right]}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3cos}\:\mathrm{3}{x}}{\mathrm{sin}\:\mathrm{3}{x}}×\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2cos}\:\mathrm{2}{x}} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{3}{x}}\:=\:\mathrm{1} \\ $$

Answered by mathmax by abdo last updated on 20/Dec/20

let f(x)=((ln(sin(3x)))/(ln(sin(2x))))  for x∼0   we have sin(3x)∼3x and sin(2x)∼2x ⇒  f(x)∼((ln(3x))/(ln(2x)))  by hospital we get lim_(x→0) f(x)=lim_(x→0) ((3/(3x))/(2/(2x))) =1

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ln}\left(\mathrm{sin}\left(\mathrm{3x}\right)\right)}{\mathrm{ln}\left(\mathrm{sin}\left(\mathrm{2x}\right)\right)}\:\:\mathrm{for}\:\mathrm{x}\sim\mathrm{0}\:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{sin}\left(\mathrm{3x}\right)\sim\mathrm{3x}\:\mathrm{and}\:\mathrm{sin}\left(\mathrm{2x}\right)\sim\mathrm{2x}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{ln}\left(\mathrm{3x}\right)}{\mathrm{ln}\left(\mathrm{2x}\right)}\:\:\mathrm{by}\:\mathrm{hospital}\:\mathrm{we}\:\mathrm{get}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \frac{\frac{\mathrm{3}}{\mathrm{3x}}}{\frac{\mathrm{2}}{\mathrm{2x}}}\:=\mathrm{1} \\ $$

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