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Question Number 82924 by jagoll last updated on 26/Feb/20

lim_(x→0) (((√(tan x))−(√(sin x)))/(x^2 (√x)))

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

Commented by jagoll last updated on 26/Feb/20

the ans : (1/4)

$$\mathrm{the}\:\mathrm{ans}\::\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by john santu last updated on 26/Feb/20

lim_(x→0) ((tan x−sin x)/(x^2 (√x) ((√(tan x)) +(√(sin x))))) = lim_(x→0) ((tan x(1−cos^2 x))/(x^2 (√x) ((√x)+(√x)))) = note lim_(x→0) tan x = lim_(x→0) sin x ≈ lim_(x→0) x lim_(x→0) ((x. ((1/2)x^2 ))/(2x^3 )) = (((((1/2)))/2)) = (1/4)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}\:\left(\sqrt{\mathrm{tan}\:\mathrm{x}}\:+\sqrt{\mathrm{sin}\:\mathrm{x}}\right)}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}\:\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}}\right)}\:= \\ $$$$\mathrm{note}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{tan}\:\mathrm{x}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{sin}\:\mathrm{x}\:\approx\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{x} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}.\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{2x}^{\mathrm{3}} }\:=\:\left(\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$

Answered by TANMAY PANACEA last updated on 26/Feb/20

lim_(x→0) (((√(sinx)) ×((1/((√(cosx)) ))−1))/(x^2 ×(√x))) lim_(x→0) (((sinx)/x))^(1/2) ×((1−cosx)/x^2 )×(1/(1+(√(cosx)))) lim_(x→0) (((sinx)/x))^(1/2) ×((2sin(x/2)×sin(x/2))/((x/2)×(x/2)×4))×(1/(1+(√(cosx)))) lim_(x→0) (((x−(x^3 /(3!))+(x^5 /(5!))+..)/x))^(1/2) ×(1/2)×(((sin(x/2))/(x/2)))^2 ×(1/(1+(√(cosx)))) =1×(1/2)×1×(1/2)=(1/4)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{{sinx}}\:×\left(\frac{\mathrm{1}}{\sqrt{{cosx}}\:}−\mathrm{1}\right)}{{x}^{\mathrm{2}} ×\sqrt{{x}}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{sinx}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ×\frac{\mathrm{1}−{cosx}}{{x}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{cosx}}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{sinx}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ×\frac{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}×{sin}\frac{{x}}{\mathrm{2}}}{\frac{{x}}{\mathrm{2}}×\frac{{x}}{\mathrm{2}}×\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{cosx}}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+..}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ×\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{{sin}\frac{{x}}{\mathrm{2}}}{\frac{{x}}{\mathrm{2}}}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{cosx}}} \\ $$$$=\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by jagoll last updated on 26/Feb/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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