Question Number 207816 by efronzo1 last updated on 27/May/24
![lim_(x→2) ((((x^2 +4))^(1/3) −(√(x^3 −4)))/( (√(x^2 −4))−((x−2))^(1/3) ))](https://www.tinkutara.com/question/Q207816.png)
$$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}−\sqrt{\mathrm{x}^{\mathrm{3}} −\mathrm{4}}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{4}}−\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}}} \\ $$
Commented by Frix last updated on 27/May/24
![I think it′s 0](https://www.tinkutara.com/question/Q207819.png)
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{0} \\ $$
Answered by Berbere last updated on 28/May/24
![x=2+y =lim_(y→0) ((((8+4y+4y^2 ))^(1/3) −(√(4+12y+6y^2 +y^3 )))/( (√(4y+y^2 ))−(y)^(1/3) )) =lim_(y→0) (((1/2)((√(1+(y/2)+(y^2 /2)))−(√(1+(y^3 /4)+((3y^2 )/2)+3y))))/( (√y)+(√(4y+y^2 )))) (√(1+(y/2)+(y^2 /2)))−(√(1+3y+(y^3 /4)+((3y^2 )/2)))=1+(1/2)((y/2))−1−(1/3)((y/2))+o(y)=(y/(12))+o(y) =lim_(y→0) ((y/(12))/( −(y)^(1/3) +(√(4y+y^2 ))))=(y^(2/3) /(12(−1+(√(4y^(1/3) +y^(4/3) )))))=0](https://www.tinkutara.com/question/Q207830.png)
$${x}=\mathrm{2}+{y} \\ $$$$=\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{\mathrm{8}+\mathrm{4}{y}+\mathrm{4}{y}^{\mathrm{2}} }−\sqrt{\mathrm{4}+\mathrm{12}{y}+\mathrm{6}{y}^{\mathrm{2}} +{y}^{\mathrm{3}} }}{\:\sqrt{\mathrm{4}{y}+{y}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{y}}} \\ $$$$=\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{1}+\frac{{y}}{\mathrm{2}}+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}}−\sqrt{\mathrm{1}+\frac{\boldsymbol{{y}}^{\mathrm{3}} }{\mathrm{4}}+\frac{\mathrm{3}\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}\boldsymbol{{y}}}\right)}{\:\sqrt{\boldsymbol{{y}}}+\sqrt{\mathrm{4}\boldsymbol{{y}}+\boldsymbol{{y}}^{\mathrm{2}} }} \\ $$$$\sqrt{\mathrm{1}+\frac{\boldsymbol{{y}}}{\mathrm{2}}+\frac{\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{2}}}−\sqrt{\mathrm{1}+\mathrm{3}\boldsymbol{{y}}+\frac{\boldsymbol{{y}}^{\mathrm{3}} }{\mathrm{4}}+\frac{\mathrm{3}\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{2}}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\boldsymbol{{y}}}{\mathrm{2}}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\boldsymbol{{y}}}{\mathrm{2}}\right)+\boldsymbol{{o}}\left(\boldsymbol{{y}}\right)=\frac{\boldsymbol{{y}}}{\mathrm{12}}+\boldsymbol{{o}}\left(\boldsymbol{{y}}\right) \\ $$$$\left.=\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{y}}{\mathrm{12}}}{\:−\sqrt[{\mathrm{3}}]{{y}}+\sqrt{\mathrm{4}{y}+{y}^{\mathrm{2}} }}=\frac{{y}^{\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{12}\left(−\mathrm{1}+\sqrt{\mathrm{4}{y}^{\frac{\mathrm{1}}{\mathrm{3}}} +{y}^{\frac{\mathrm{4}}{\mathrm{3}}} }\right.}\right)=\mathrm{0} \\ $$$$ \\ $$