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Question Number 115318 by bemath last updated on 25/Sep/20

 lim_(x→0)  ((xsin x)/(2sin^2 (3x)−x^2 cos x))

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\mathrm{sin}\:{x}}{\mathrm{2sin}\:^{\mathrm{2}} \left(\mathrm{3}{x}\right)−{x}^{\mathrm{2}} \mathrm{cos}\:{x}} \\ $$

Answered by bobhans last updated on 25/Sep/20

lim_(x→0)  ((x sin x)/(2 sin^2 (3x)−x^2 cos x)) =  lim_(x→0)  ((x^2  (((sin x)/x)))/(2x^2 (((sin 3x)/x))^2 −x^2  cos x)) =  lim_(x→0)  (x^2 /(x^2 (2.9−1))) = (1/(17))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\mathrm{sin}\:{x}}{\mathrm{2}\:\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{3}{x}\right)−{x}^{\mathrm{2}} \mathrm{cos}\:{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \:\left(\frac{\mathrm{sin}\:{x}}{{x}}\right)}{\mathrm{2}{x}^{\mathrm{2}} \left(\frac{\mathrm{sin}\:\mathrm{3}{x}}{{x}}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \:\mathrm{cos}\:{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left(\mathrm{2}.\mathrm{9}−\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{17}} \\ $$

Answered by Bird last updated on 25/Sep/20

let f(x)=((xsinx)/(2sin^2 (3x)−x^2 cosx))  we have xsinx∼x^2   sin^2 (3x)∼9x^2   x^2 cosx ∼x^2 (1−(x^2 /2)) ⇒  f(x)∼(x^2 /(18x^2 −x^2  +(x^4 /2))) =(x^2 /(17x^2 +(x^4 /2)))  f(x)∼(1/(17+(x^2 /2))) ⇒lim_(x→0)   f(x)=(1/(17))

$${let}\:{f}\left({x}\right)=\frac{{xsinx}}{\mathrm{2}{sin}^{\mathrm{2}} \left(\mathrm{3}{x}\right)−{x}^{\mathrm{2}} {cosx}} \\ $$$${we}\:{have}\:{xsinx}\sim{x}^{\mathrm{2}} \\ $$$${sin}^{\mathrm{2}} \left(\mathrm{3}{x}\right)\sim\mathrm{9}{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {cosx}\:\sim{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\sim\frac{{x}^{\mathrm{2}} }{\mathrm{18}{x}^{\mathrm{2}} −{x}^{\mathrm{2}} \:+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}\:=\frac{{x}^{\mathrm{2}} }{\mathrm{17}{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{4}} }{\mathrm{2}}} \\ $$$${f}\left({x}\right)\sim\frac{\mathrm{1}}{\mathrm{17}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{17}} \\ $$

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