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Question Number 82950 by jagoll last updated on 26/Feb/20
lim_(x→0) ((((1+tan x))^(1/(3 )) −((1+sin x))^(1/(3 )) )/x^3 ) =
$$ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}\:−\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\mathrm{3}} }\:=\: \\ $$
Commented by mathmax by abdo last updated on 26/Feb/20
we have a^3 −b^3 =(a−b)(a^2 +ab +b^2 ) ⇒ a−b =(^3 (√a)−^3 (√b))( (^3 (√a))^2 +^3 (√(ab))+(^3 (√b))^2 ) ⇒ ⇒^3 (1/x^3 )((√(1+tanx))−^3 (√(1+sinx)))=(1/x^3 )×(((1+tanx−1−sinx)/((1+tanx)^(2/3) +(1+tanx)(1+sinx)^(1/3) +(1+sinx)^(2/3) )))⇒ lim_(x→0) (...)=lim_(x→0) (1/(3x^3 ))(((sinx)/(cosx))−sinx) =(1/3)lim_(x→0) ((sinx)/x) ×((1−cosx)/(x^2 ×cosx)) =(1/3)×1×(1/2) =(1/6)
$${we}\:{have}\:{a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\left({a}−{b}\right)\left({a}^{\mathrm{2}} \:+{ab}\:+{b}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${a}−{b}\:=\left(^{\mathrm{3}} \sqrt{{a}}−^{\mathrm{3}} \sqrt{{b}}\right)\left(\:\left(^{\mathrm{3}} \sqrt{{a}}\right)^{\mathrm{2}} \:+^{\mathrm{3}} \sqrt{{ab}}+\left(^{\mathrm{3}} \sqrt{{b}}\right)^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\Rightarrow^{\mathrm{3}} \frac{\mathrm{1}}{{x}^{\mathrm{3}} }\left(\sqrt{\mathrm{1}+{tanx}}−^{\mathrm{3}} \sqrt{\mathrm{1}+{sinx}}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{3}} }×\left(\frac{\mathrm{1}+{tanx}−\mathrm{1}−{sinx}}{\left(\mathrm{1}+{tanx}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\left(\mathrm{1}+{tanx}\right)\left(\mathrm{1}+{sinx}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:+\left(\mathrm{1}+{sinx}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }\right)\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left(…\right)={lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }\left(\frac{{sinx}}{{cosx}}−{sinx}\right)\:=\frac{\mathrm{1}}{\mathrm{3}}{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{sinx}}{{x}}\:×\frac{\mathrm{1}−{cosx}}{{x}^{\mathrm{2}} ×{cosx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Answered by john santu last updated on 26/Feb/20
lim_(x→0) ((tan x(1−cos x))/x^3 ) × lim_(x→0) (1/( (((1+tan x)^2 ))^(1/(3 )) +(((1+sin x)^2 ))^(1/(3 )) +(((1+tan x)(1+sin x)))^(1/(3 )) )) = (1/2) × (1/3) = (1/6)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\mathrm{x}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{x}^{\mathrm{3}} }\:×\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}\:}]{\left(\mathrm{1}+\mathrm{tan}\:\mathrm{x}\right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}\:}]{\left(\mathrm{1}+\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}\:}]{\left(\mathrm{1}+\mathrm{tan}\:\mathrm{x}\right)\left(\mathrm{1}+\mathrm{sin}\:\mathrm{x}\right)}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:×\:\frac{\mathrm{1}}{\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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