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Question Number 184792 by Spillover last updated on 11/Jan/23

Evaluate   lim_(x→0)  ((tan x−sin x)/(sin^3 x))

$${Evaluate}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}−\mathrm{sin}\:{x}}{\mathrm{sin}\:^{\mathrm{3}} {x}} \\ $$$$ \\ $$

Commented by MJS_new last updated on 11/Jan/23

(1/2)

$$\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by Spillover last updated on 11/Jan/23

thank you

$${thank}\:{you} \\ $$

Answered by aba last updated on 12/Jan/23

let x=2u  lim_(x→0) ((tan(x)−sin(x))/(sin^3 (x)))=lim_(u→0) ((tan(2u)−sin(2u))/(sin^3 (2u)))                                         =lim_(u→0) ((sin(2u)((1/(cos(2u)))−1))/(sin^3 (2u)))                                         =lim_(u→0) (((1/(cos(2u)))−1)/(sin^2 (2u)))                                         =lim_(u→0) ((1−cos(2u))/(cos(2u)(1−cos^2 (u))))                                        =lim_(u→0) (1/(cos(2u)(1+cos(2u))))                                       =(1/2)

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

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