lim-x-0-x-sin-x-x-2-x-

Question Number 137902 by benjo_mathlover last updated on 08/Apr/21

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

Answered by greg_ed last updated on 08/Apr/21

L′Hopital′s rule lim_(x→0) ((x−sin x)/(x^2 (√x))) = lim_(x→0) (((d/dx)(x−sin x))/((d/dx)(x^2 (√x)))) = lim_(x→0) ((1−cos x)/((5x(√x))/2)) = lim_(x→0) ((2−2cos x)/(5x(√x))) = lim_(x→0) (((d/dx)(2−2cos x))/((d/dx)(5x(√x)))) = lim_(x→0) ((2sin x)/((15(√x))/2)) = lim_(x→0) ((4sin x)/(15(√x))) = lim_(x→0) (((d/dx)(4sin x))/((d/dx)(15(√x)))) = lim_(x→0) ((4cos x)/((15)/(2(√x)))) = lim_(x→0) ((8(√x) cos x)/(15)) plug in x = 0, and we get : lim_(x→0) ((x−sin x)/(x^2 (√x))) = 0.

$$ \\ $$$$\boldsymbol{\mathrm{L}}'\boldsymbol{\mathrm{Hopital}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{rule}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−{sin}\:{x}}{{x}^{\mathrm{2}} \sqrt{{x}}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left({x}−{sin}\:{x}\right)}{\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} \sqrt{{x}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−{cos}\:{x}}{\frac{\mathrm{5}{x}\sqrt{{x}}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}−\mathrm{2}{cos}\:{x}}{\mathrm{5}{x}\sqrt{{x}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left(\mathrm{2}−\mathrm{2}{cos}\:{x}\right)}{\frac{{d}}{{dx}}\left(\mathrm{5}{x}\sqrt{{x}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{sin}\:{x}}{\frac{\mathrm{15}\sqrt{{x}}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4}{sin}\:{x}}{\mathrm{15}\sqrt{{x}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left(\mathrm{4}{sin}\:{x}\right)}{\frac{{d}}{{dx}}\left(\mathrm{15}\sqrt{{x}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4}{cos}\:{x}}{\frac{\mathrm{15}}{\mathrm{2}\sqrt{{x}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{8}\sqrt{{x}}\:{cos}\:{x}}{\mathrm{15}} \\ $$$$\boldsymbol{\mathrm{plug}}\:\boldsymbol{\mathrm{in}}\:{x}\:=\:\mathrm{0},\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}}\::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\frac{\boldsymbol{{x}}−\boldsymbol{{sin}}\:\boldsymbol{{x}}}{\boldsymbol{{x}}^{\mathrm{2}} \sqrt{\boldsymbol{{x}}}}\:=\:\mathrm{0}. \\ $$

Answered by EnterUsername last updated on 08/Apr/21

lim_(x→0) ((x−sinx)/(x^2 (√x))) =lim_(x→0) ((x−(x−(x^3 /3)))/(x^2 (√x)))=lim_(x→0) (x^3 /(3x^2 (√x))) =lim_(x→0) (x^(3−(5/2)) /3)=lim_(x→0) ((√x)/3)=0

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}−{sinx}}{{x}^{\mathrm{2}} \sqrt{{x}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)}{{x}^{\mathrm{2}} \sqrt{{x}}}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}^{\mathrm{3}} }{\mathrm{3}{x}^{\mathrm{2}} \sqrt{{x}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}^{\mathrm{3}−\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{3}}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\sqrt{{x}}}{\mathrm{3}}=\mathrm{0} \\ $$

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