Question Number 195982 by mathlove last updated on 14/Aug/23
![lim_(x→1) [(1/(2(1−(√x))))−(1/(3(1−(x)^(1/3) )))]=? with out l′pital rule](https://www.tinkutara.com/question/Q195982.png)
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\sqrt{{x}}\right)}−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{{x}}\right)}\right]=? \\ $$$${with}\:{out}\:{l}'{pital}\:{rule} \\ $$
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=(x)^(1/6) L=lim_(t→1) [(1/(2(1−t^3 )))−(1/(3(1−t^2 )))] =lim_(t→1) [((1+t^3 )/(2(1−t^6 )))−((1+t^2 +t^4 )/(3(1−t^6 )))] =lim_(t→1) [((1+2t^3 −2t^2 −2t^4 )/(6(1−t^6 )))] =lim_(t→1) [((1−t^3 −2t^2 +4t^3 −2t^4 )/(6(1−t^6 )))] =lim_(t→1) [((1−t^3 −2t^2 (1−t)^2 )/(6(1−t^6 )))] =lim_(t→1) [(1/(6(1+t^3 )))−((t^2 (1−t)^2 )/(3(1+t^3 )(1−t^3 )))] =lim_(t→1) [(1/(6(1+t^3 )))−((t^2 (1−t))/(3(1+t^3 )(1+t+t^2 )))] =(1/(6(1+1)))−((1×(1−1))/(3(1+1)(1+1+1))) =(1/(12))](https://www.tinkutara.com/question/Q195985.png)
$${let}\:{t}=\sqrt[{\mathrm{6}}]{{x}} \\ $$$${L}=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{3}} \right)}−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}+{t}^{\mathrm{3}} }{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{6}} \right)}−\frac{\mathrm{1}+{t}^{\mathrm{2}} +{t}^{\mathrm{4}} }{\mathrm{3}\left(\mathrm{1}−{t}^{\mathrm{6}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}+\mathrm{2}{t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{4}} }{\mathrm{6}\left(\mathrm{1}−{t}^{\mathrm{6}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}−{t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}{t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{4}} }{\mathrm{6}\left(\mathrm{1}−{t}^{\mathrm{6}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}−{t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{\mathrm{6}\left(\mathrm{1}−{t}^{\mathrm{6}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{6}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)}−\frac{{t}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{\mathrm{3}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)\left(\mathrm{1}−{t}^{\mathrm{3}} \right)}\right] \\ $$$$\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{6}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)}−\frac{{t}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{\mathrm{3}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)}\right] \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{6}\left(\mathrm{1}+\mathrm{1}\right)}−\frac{\mathrm{1}×\left(\mathrm{1}−\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{1}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}+\mathrm{1}\right)} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{12}} \\ $$
Commented by AROUNAMoussa last updated on 14/Aug/23
$$\boldsymbol{{cool}}! \\ $$
Commented by mathlove last updated on 15/Aug/23
$${thanks} \\ $$