Question Number 195982 by mathlove last updated on 14/Aug/23 | ||
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\sqrt{{x}}\right)}−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{{x}}\right)}\right]=? \\ $$$${with}\:{out}\:{l}'{pital}\:{rule} \\ $$ | ||
Answered by MM42 last updated on 14/Aug/23 | ||
$${lim}_{{x}\rightarrow\mathrm{1}} \:\left(\frac{\mathrm{1}+\sqrt{{x}}}{\mathrm{2}\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}{\mathrm{3}\left(\mathrm{1}−{x}\right)}\right) \\ $$$$={lim}_{{x}\rightarrow\mathrm{1}} \:\left(\:\frac{\mathrm{1}+\mathrm{3}\sqrt[{\mathrm{6}}]{{x}^{\mathrm{3}} }−\mathrm{2}\sqrt[{\mathrm{6}}]{{x}^{\mathrm{2}} }−\mathrm{2}\sqrt[{\mathrm{6}}]{{x}^{\mathrm{4}} }}{\mathrm{6}\left(\mathrm{1}−{x}\right)}\:\:\right) \\ $$$$={lim}_{{x}\rightarrow\mathrm{1}} \:\left(\:\frac{\left(\mathrm{1}−\sqrt[{\mathrm{6}}]{{x}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{6}}]{{x}}−\sqrt[{\mathrm{6}}]{{x}^{\mathrm{2}} }+\mathrm{2}\sqrt[{\mathrm{6}}]{{x}^{\mathrm{3}} \:}\right)}{\mathrm{6}\left(\mathrm{1}−\sqrt[{\mathrm{6}}]{{x}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{6}}]{{x}}+\sqrt[{\mathrm{6}}]{{x}^{\mathrm{2}} }+\sqrt[{\mathrm{6}}]{{x}^{\mathrm{3}} }+\sqrt[{\mathrm{6}}]{{x}^{\mathrm{4}} }+\sqrt[{\mathrm{6}}]{{x}^{\mathrm{5}} \:}\right)}\:\:\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\:\checkmark \\ $$$$ \\ $$ | ||
Commented by mathlove last updated on 15/Aug/23 | ||
$${thanks} \\ $$ | ||
Answered by mr W last updated on 14/Aug/23 | ||
$${let}\:{t}=\sqrt[{\mathrm{6}}]{{x}} \\ $$$${L}=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{3}} \right)}−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}+{t}^{\mathrm{3}} }{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{6}} \right)}−\frac{\mathrm{1}+{t}^{\mathrm{2}} +{t}^{\mathrm{4}} }{\mathrm{3}\left(\mathrm{1}−{t}^{\mathrm{6}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}+\mathrm{2}{t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{4}} }{\mathrm{6}\left(\mathrm{1}−{t}^{\mathrm{6}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}−{t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}{t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{4}} }{\mathrm{6}\left(\mathrm{1}−{t}^{\mathrm{6}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}−{t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{\mathrm{6}\left(\mathrm{1}−{t}^{\mathrm{6}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{6}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)}−\frac{{t}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{\mathrm{3}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)\left(\mathrm{1}−{t}^{\mathrm{3}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{6}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)}−\frac{{t}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{\mathrm{3}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)}\right] \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{6}\left(\mathrm{1}+\mathrm{1}\right)}−\frac{\mathrm{1}×\left(\mathrm{1}−\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{1}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}+\mathrm{1}\right)} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{12}} \\ $$ | ||
Commented by AROUNAMoussa last updated on 14/Aug/23 | ||
$$\boldsymbol{{cool}}! \\ $$ | ||
Commented by mathlove last updated on 15/Aug/23 | ||
$${thanks} \\ $$ | ||