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Question Number 121302 by liberty last updated on 06/Nov/20

(1)lim_(x→0) (((1−e^(2x) )sin (3x))/(∣4x∣)) ? (2) lim_(x→0) ⌊ (((1−e^(2x) )sin 3x)/(∣4x∣)) ⌋ ?

$$\:\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{e}^{\mathrm{2x}} \right)\mathrm{sin}\:\left(\mathrm{3x}\right)}{\mid\mathrm{4x}\mid}\:?\: \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\lfloor\:\frac{\left(\mathrm{1}−\mathrm{e}^{\mathrm{2x}} \right)\mathrm{sin}\:\mathrm{3x}}{\mid\mathrm{4x}\mid}\:\rfloor\:? \\ $$

Answered by bemath last updated on 06/Nov/20

(1) lim_(x→0) (((1−e^(2x) )sin 3x)/(∣4x∣ )) lim_(x→0^− ) (((1−e^(2x) )sin 3x)/(−4x)) = lim_(x→0^− ) ((3cos 3x(1−e^(2x) )−2e^(2x) sin 3x)/(−4)) = 0 lim_(x→0^+ ) (((1−e^(2x) )sin 3x)/(4x)) = lim_(x→0^+ ) ((3cos 3x(1−e^(2x) )−2e^(2x) sin 3x)/4) = 0

$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{e}^{\mathrm{2x}} \right)\mathrm{sin}\:\mathrm{3x}}{\mid\mathrm{4x}\mid\:} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{e}^{\mathrm{2x}} \right)\mathrm{sin}\:\mathrm{3x}}{−\mathrm{4x}}\:=\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\frac{\mathrm{3cos}\:\mathrm{3x}\left(\mathrm{1}−\mathrm{e}^{\mathrm{2x}} \right)−\mathrm{2e}^{\mathrm{2x}} \:\mathrm{sin}\:\mathrm{3x}}{−\mathrm{4}}\:=\:\mathrm{0}\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{e}^{\mathrm{2x}} \right)\mathrm{sin}\:\mathrm{3x}}{\mathrm{4x}}\:=\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{3cos}\:\mathrm{3x}\left(\mathrm{1}−\mathrm{e}^{\mathrm{2x}} \right)−\mathrm{2e}^{\mathrm{2x}} \:\mathrm{sin}\:\mathrm{3x}}{\mathrm{4}}\:=\:\mathrm{0} \\ $$

Answered by Bird last updated on 06/Nov/20

let f(x)=(((1−e^(2x) )sin(3x))/(4∣x∣)) we have e^(2x) ∼1+2x and sin(3x)∼3x ⇒f(x)∼((−2x(3x))/(4∣x∣))=((−6x^2 )/(4∣x∣)) =−(3/2)∣x∣ ⇒lim_(x→0) f(x)=0

$${let}\:{f}\left({x}\right)=\frac{\left(\mathrm{1}−{e}^{\mathrm{2}{x}} \right){sin}\left(\mathrm{3}{x}\right)}{\mathrm{4}\mid{x}\mid} \\ $$$${we}\:{have}\:{e}^{\mathrm{2}{x}} \:\sim\mathrm{1}+\mathrm{2}{x}\:\:{and}\:{sin}\left(\mathrm{3}{x}\right)\sim\mathrm{3}{x} \\ $$$$\Rightarrow{f}\left({x}\right)\sim\frac{−\mathrm{2}{x}\left(\mathrm{3}{x}\right)}{\mathrm{4}\mid{x}\mid}=\frac{−\mathrm{6}{x}^{\mathrm{2}} }{\mathrm{4}\mid{x}\mid} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\mid{x}\mid\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\mathrm{0} \\ $$

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