Question-186983

Question Number 186983 by cortano12 last updated on 12/Feb/23

Answered by CElcedricjunior last updated on 12/Feb/23

lim_(x→0) ((x((x−1))^(1/3) +(((x+1)))^(1/4) −1)/(x^2 ((x+1))^(1/3) +((x+1))^(1/4) −1))=k k=lim_(x→0) ((x(−1−(1/3)x+(1/9)x^2 )+(1+(1/4)x−(3/(32))x^2 )−1)/(x^2 (1+(1/3)x−(( 1)/9)x^2 )+(1+(1/4)x−(3/(32))x^2 )−1))★celebre cedric junior =lim_(x→0) ((−x−(1/3)x^2 +(1/9)x^3 +(1/4)x−(3/(32))x^2 )/(x^2 +(1/3)x^3 −(1/9)x^4 +(1/4)x−(3/(32)))) =(((1/4)−1)/(1/4))=−3 lim_(x→0) ((x((x−1))^(1/3) +((x+1))^(1/4) −1)/(x^2 ((x+1))^(1/3) +((x+1))^(1/4) −1))=−3 ★Moivre

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{x}}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}−\mathrm{1}}+\sqrt[{\mathrm{4}}]{\left(\boldsymbol{{x}}+\mathrm{1}\right)}−\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\boldsymbol{{x}}+\mathrm{1}}+\sqrt[{\mathrm{4}}]{\boldsymbol{{x}}+\mathrm{1}}−\mathrm{1}}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{k}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{x}}\left(−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{x}}+\frac{\mathrm{1}}{\mathrm{9}}\boldsymbol{{x}}^{\mathrm{2}} \right)+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{x}}−\frac{\mathrm{3}}{\mathrm{32}}\boldsymbol{{x}}^{\mathrm{2}} \right)−\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{x}}−\frac{\:\:\mathrm{1}}{\mathrm{9}}\boldsymbol{{x}}^{\mathrm{2}} \right)+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{x}}−\frac{\mathrm{3}}{\mathrm{32}}\boldsymbol{{x}}^{\mathrm{2}} \right)−\mathrm{1}}\bigstar{celebre}\:{cedric}\:{junior} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\boldsymbol{{x}}−\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{x}}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{9}}\boldsymbol{{x}}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{x}}−\frac{\mathrm{3}}{\mathrm{32}}\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{{x}}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{x}}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{9}}\boldsymbol{{x}}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{x}}−\frac{\mathrm{3}}{\mathrm{32}}} \\ $$$$=\frac{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{4}}}=−\mathrm{3} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{x}}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}−\mathrm{1}}+\sqrt[{\mathrm{4}}]{\boldsymbol{{x}}+\mathrm{1}}−\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\boldsymbol{{x}}+\mathrm{1}}+\sqrt[{\mathrm{4}}]{\boldsymbol{{x}}+\mathrm{1}}−\mathrm{1}}=−\mathrm{3}\:\:\:\bigstar{Moivre} \\ $$

Answered by ARUNG_Brandon_MBU last updated on 12/Feb/23

L=lim_(x→0) ((x((x−1))^(1/3) +((x+1))^(1/4) −1)/(x^2 ((x+1))^(1/3) +((x+1))^(1/4) −1)) =lim_(x→0) ((−x(1−(x/3))+(1+(x/4))−1)/(x^2 (1+(x/3))+(1+(x/4))−1)) =lim_(x→0) (((x^2 /3)−((3x)/4))/((x^3 /3)+x^2 +(x/4)))=lim_(x→0) (((x/3)−(3/4))/((x^2 /3)+x+(1/4)))=−3

$$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}+\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}−\mathrm{1}}{{x}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}+\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}−\mathrm{1}} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−{x}\left(\mathrm{1}−\frac{{x}}{\mathrm{3}}\right)+\left(\mathrm{1}+\frac{{x}}{\mathrm{4}}\right)−\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{{x}}{\mathrm{3}}\right)+\left(\mathrm{1}+\frac{{x}}{\mathrm{4}}\right)−\mathrm{1}} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{3}{x}}{\mathrm{4}}}{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{x}^{\mathrm{2}} +\frac{{x}}{\mathrm{4}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{x}}{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{4}}}{\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+{x}+\frac{\mathrm{1}}{\mathrm{4}}}=−\mathrm{3} \\ $$

Answered by cortano12 last updated on 14/Feb/23

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