Question-184620

Question Number 184620 by Ml last updated on 09/Jan/23

Commented by Ml last updated on 09/Jan/23

$$\mathrm{please}\:\mathrm{solution}???? \\ $$

Answered by mr W last updated on 09/Jan/23

=lim_(x→0) ((sin x^3 )/x^3 )×((tan x)/x)×(((sin (x/2))/(x/2)))^2 ×((x/(sin x)))^5 ×(x^(1/6) /(2(x+x^(1/3) )^(1/2) )) =lim_(x→0) (x^(1/6) /(2(x+x^(1/3) )^(1/2) )) =lim_(x→0) (1/(2(x^(4/3) +1)^(1/2) )) =(1/2)

$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} }×\frac{\mathrm{tan}\:{x}}{{x}}×\left(\frac{\mathrm{sin}\:\frac{{x}}{\mathrm{2}}}{\frac{{x}}{\mathrm{2}}}\right)^{\mathrm{2}} ×\left(\frac{{x}}{\mathrm{sin}\:{x}}\right)^{\mathrm{5}} ×\frac{{x}^{\frac{\mathrm{1}}{\mathrm{6}}} }{\mathrm{2}\left({x}+{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\frac{\mathrm{1}}{\mathrm{6}}} }{\mathrm{2}\left({x}+{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}\left({x}^{\frac{\mathrm{4}}{\mathrm{3}}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by qaz last updated on 10/Jan/23

lim_(x→0^+ ) ((sin x^3 tan x(1−cos x))/( (√(x+(x)^(1/3) ))((x^5 )^(1/6) sin^5 x))) =lim_(x→0^+ ) ((x^3 ∙x∙(1/2)x^2 )/( ((x)^(1/3) )^(1/2) ∙(x^5 )^(1/6) ∙x^5 )) =(1/2) −−−−−−−−−−− x→0^+ ,sin x^3 ∼x^3 tan x∼x 1−cos x∼(1/2)x^2 (√(x+(x)^(1/3) ))∼((x)^(1/3) )^(1/2) sin^5 x∼x^5

$$\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{\mathrm{sin}\:{x}^{\mathrm{3}} \mathrm{tan}\:{x}\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{\:\sqrt{{x}+\sqrt[{\mathrm{3}}]{{x}}}\left(\sqrt[{\mathrm{6}}]{{x}^{\mathrm{5}} }\mathrm{sin}\:^{\mathrm{5}} {x}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{{x}^{\mathrm{3}} \centerdot{x}\centerdot\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} }{\:\sqrt[{\mathrm{2}}]{\sqrt[{\mathrm{3}}]{{x}}}\centerdot\sqrt[{\mathrm{6}}]{{x}^{\mathrm{5}} }\centerdot{x}^{\mathrm{5}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$−−−−−−−−−−− \\ $$$${x}\rightarrow\mathrm{0}^{+} ,{sin}\:{x}^{\mathrm{3}} \sim{x}^{\mathrm{3}} \:\:\:\:{tan}\:{x}\sim{x}\:\:\:\mathrm{1}−\mathrm{cos}\:{x}\sim\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \\ $$$$\sqrt{{x}+\sqrt[{\mathrm{3}}]{{x}}}\sim\sqrt[{\mathrm{2}}]{\sqrt[{\mathrm{3}}]{{x}}}\:\:\:\:\:\:\:\:\mathrm{sin}\:^{\mathrm{5}} {x}\sim{x}^{\mathrm{5}} \\ $$

Commented by Ml last updated on 09/Jan/23

$$\mathrm{step}\:\mathrm{by}\:\mathrm{step}\:???? \\ $$

Answered by cortano1 last updated on 10/Jan/23

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