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Question Number 149251 by john_santu last updated on 04/Aug/21

  lim_(x→0^+ ) ((tan x (√(tan x))−sin x (√(sin x)))/(x^3  (√x))) =?

$$\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{tan}\:\mathrm{x}\:\sqrt{\mathrm{tan}\:\mathrm{x}}−\mathrm{sin}\:\mathrm{x}\:\sqrt{\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\mathrm{3}} \:\sqrt{\mathrm{x}}}\:=? \\ $$$$ \\ $$

Answered by dumitrel last updated on 04/Aug/21

lim_(x→0^+ ) ((sinx)/x)∙lim_(x→0^0 ) (√((sinx)/x))∙lim_(x→0^+ ) (1/x^2 )((1/(cosx(√(cosx))))−1)=  lim_(x→0^+ ) ((1−cos^3 x)/(x^2 cosx(√(cosx))(1+cosx(√(cosx)))))=lim_(x→0^+ ) (((1−cosx)(1+cosx+cos^2 x))/(x^2 cosx(√(cosx))(1+cosx(√(cosx)))))=  =lim_(x→0^+ ) ((2sin^2 (x/2)(1+cosx+cos^2 x))/(x^2 cosx(√(cosx))(1+cosx(√(cosx)))))=  =lim_(x→0^+ ) (((sin(x/2))/(x/2)))^2 (((1+cosx+cos^2 x))/(2cosx(√(cosx))(1+cosx(√(cosx)))))=(3/4)

$$\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{{sinx}}{{x}}\centerdot\underset{{x}\rightarrow\mathrm{0}^{\mathrm{0}} } {{lim}}\sqrt{\frac{{sinx}}{{x}}}\centerdot\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{{cosx}\sqrt{{cosx}}}−\mathrm{1}\right)= \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{\mathrm{1}−{cos}^{\mathrm{3}} {x}}{{x}^{\mathrm{2}} {cosx}\sqrt{{cosx}}\left(\mathrm{1}+{cosx}\sqrt{{cosx}}\right)}=\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{\left(\mathrm{1}−{cosx}\right)\left(\mathrm{1}+{cosx}+{cos}^{\mathrm{2}} {x}\right)}{{x}^{\mathrm{2}} {cosx}\sqrt{{cosx}}\left(\mathrm{1}+{cosx}\sqrt{{cosx}}\right)}= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\left(\mathrm{1}+{cosx}+{cos}^{\mathrm{2}} {x}\right)}{{x}^{\mathrm{2}} {cosx}\sqrt{{cosx}}\left(\mathrm{1}+{cosx}\sqrt{{cosx}}\right)}= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\left(\frac{{sin}\frac{{x}}{\mathrm{2}}}{\frac{{x}}{\mathrm{2}}}\right)^{\mathrm{2}} \frac{\left(\mathrm{1}+{cosx}+{cos}^{\mathrm{2}} {x}\right)}{\mathrm{2}{cosx}\sqrt{{cosx}}\left(\mathrm{1}+{cosx}\sqrt{{cosx}}\right)}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$

Answered by john_santu last updated on 04/Aug/21

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