Question Number 198123 by a.lgnaoui last updated on 10/Oct/23
$$\mathrm{Determiner} \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{3}} \:\frac{\boldsymbol{\mathrm{x}}−\mathrm{3}}{\:^{\mathrm{3}} \sqrt{\boldsymbol{\mathrm{x}}+\mathrm{5}}\:−\mathrm{2}} \\ $$$$ \\ $$
Answered by Mathspace last updated on 10/Oct/23
$${a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${a}−{b}=\left(^{\mathrm{3}} \sqrt{{a}}−^{\mathrm{3}} \sqrt{{b}}\right)\left(\left(^{\mathrm{3}} \sqrt{{a}}\right)^{\mathrm{2}} +^{\mathrm{3}} \sqrt{{ab}}+\left(^{\mathrm{3}} \sqrt{{b}}\right)^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\left(^{\mathrm{3}} \sqrt{{x}+\mathrm{5}}\right)−\left(^{\mathrm{3}} \sqrt{\mathrm{8}}\right) \\ $$$$=\frac{{x}+\mathrm{5}−\mathrm{8}}{\left({x}+\mathrm{5}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\left(\mathrm{8}\left({x}+\mathrm{5}\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{8}^{\frac{\mathrm{2}}{\mathrm{3}}} } \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{3}} {f}\left({x}\right) \\ $$$$={lim}_{{x}\rightarrow\mathrm{3}} \frac{{x}−\mathrm{3}}{\frac{{x}−\mathrm{3}}{\left({x}+\mathrm{5}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{2}\left({x}+\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{4}}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{3}} \:\left({x}+\mathrm{5}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{2}\left({x}+\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{4} \\ $$$$=\mathrm{8}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{2}×\mathrm{8}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{4} \\ $$$$=\mathrm{4}+\mathrm{4}+\mathrm{4}=\mathrm{12} \\ $$
Commented by a.lgnaoui last updated on 10/Oct/23
$$\mathrm{exact}\:\: \\ $$
Answered by MM42 last updated on 10/Oct/23
$$\sqrt[{\mathrm{3}}]{{x}+\mathrm{5}}−\mathrm{2}={u}\Rightarrow{x}+\mathrm{5}=\left({u}+\mathrm{2}\right)^{\mathrm{3}} \Rightarrow{x}=\left({u}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{5} \\ $$$$\Rightarrow{lim}_{{u}\rightarrow\mathrm{0}} \:\frac{\left({u}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{8}}{{u}}={lim}_{{u}\rightarrow\mathrm{0}} \:\frac{{u}\left({u}^{\mathrm{2}} +\mathrm{6}{u}+\mathrm{12}\right)}{{u}}=\mathrm{12}\:\checkmark \\ $$
Answered by cortano12 last updated on 11/Oct/23
$$\:\:\underline{\underbrace{\:}} \underbrace{\mathcal{Y}}\mathrm{3} \\ $$