Question-167699

Question Number 167699 by HongKing last updated on 23/Mar/22

Answered by alephzero last updated on 23/Mar/22

1. lim_(x→0) ((1−cos x)/(x sin x)) =^(use L′Ho^� pital′s rule) = lim ((sin x)/(sin x+x cos x)) =^(again) = lim ((cos x)/(cos x−sin x+cos x)) = = lim ((cos x)/(2cos x−sin x)) = (1/(2×1−0)) = (1/2)

$$\mathrm{1}.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}\:\mathrm{sin}\:{x}}\:\overset{\mathrm{use}\:\mathrm{L}'\mathrm{H}\hat {\mathrm{o}pital}'\mathrm{s}\:\mathrm{rule}} {=} \\ $$$$=\:\mathrm{lim}\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}+{x}\:\mathrm{cos}\:{x}}\:\overset{\mathrm{again}} {=} \\ $$$$=\:\mathrm{lim}\:\frac{\mathrm{cos}\:{x}}{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:= \\ $$$$=\:\mathrm{lim}\:\frac{\mathrm{cos}\:{x}}{\mathrm{2cos}\:{x}−\mathrm{sin}\:{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{1}−\mathrm{0}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by alephzero last updated on 23/Mar/22

2. L′Hopital again lim((3cos^2 2x−cos 2x−2)/(x cos x)) = lim((−6sin 4x+2sin 2x)/(cos x−x sin x)) = ((0+0)/(1−0)) = 0

$$\mathrm{2}.\:\mathrm{L}'\mathrm{Hopital}\:\mathrm{again} \\ $$$$\mathrm{lim}\frac{\mathrm{3cos}^{\mathrm{2}} \:\mathrm{2}{x}−\mathrm{cos}\:\mathrm{2}{x}−\mathrm{2}}{{x}\:\mathrm{cos}\:{x}}\:=\: \\ $$$$\mathrm{lim}\frac{−\mathrm{6sin}\:\mathrm{4}{x}+\mathrm{2sin}\:\mathrm{2}{x}}{\mathrm{cos}\:{x}−{x}\:\mathrm{sin}\:{x}}\:= \\ $$$$\frac{\mathrm{0}+\mathrm{0}}{\mathrm{1}−\mathrm{0}}\:=\:\mathrm{0} \\ $$

Answered by qaz last updated on 23/Mar/22

lim_(x→0) (((e^(−3x+2) −e^2 )sin πx)/(4x^2 )) =lim_(x→0) ((e^2 πx(e^(−3x) −1))/(4x^2 )) =lim_(x→0) ((πe^2 )/(4x))(−3x+o(x)) =−(3/4)πe^2

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

Answered by malwan last updated on 24/Mar/22

1.lim_(x→0) ((1−cos x)/(x sinx)) = lim_(x→0) ((2sin^2 (x/2))/(x sin x)) =2×((1/2))^2 = (1/2)

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

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