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Question Number 148654 by mathdanisur last updated on 29/Jul/21
lim_(x→1) (((1+x)/(2+x)))^((1−(√x))/(1−x)) = ?
$$\underset{{x}\rightarrow\mathrm{1}} {{lim}}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}+{x}}\right)^{\frac{\mathrm{1}−\sqrt{\boldsymbol{{x}}}}{\mathrm{1}−\boldsymbol{{x}}}} \:=\:? \\ $$
Answered by ArielVyny last updated on 29/Jul/21
e^(((1−(√x))/(1−x))ln(((1+x)/(2+x)))) posons t=x−1 { ((x→1)),((t→0)) :} e^((((√(t+1))−1)/t)ln(((t+2)/(t+3)))) =_(t→0) e^((((√(t+1))−1)/t)ln((2/3))) ∼e^(ln((2/3))) =(2/3) (((√(t+1))−1)/t)∼_0 (t/t)∼1 conclusion lim_(x→1) (((1+x)/(2+x)))^((1−(√x))/(1−x)) =(2/3)
$${e}^{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}−{x}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}+{x}}\right)} \:\:\:\:{posons}\:{t}={x}−\mathrm{1}\:\begin{cases}{{x}\rightarrow\mathrm{1}}\\{{t}\rightarrow\mathrm{0}}\end{cases} \\ $$$${e}^{\frac{\sqrt{{t}+\mathrm{1}}−\mathrm{1}}{{t}}{ln}\left(\frac{{t}+\mathrm{2}}{{t}+\mathrm{3}}\right)} =_{{t}\rightarrow\mathrm{0}} {e}^{\frac{\sqrt{{t}+\mathrm{1}}−\mathrm{1}}{{t}}{ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)} \sim{e}^{{ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\frac{\sqrt{{t}+\mathrm{1}}−\mathrm{1}}{{t}}\underset{\mathrm{0}} {\sim}\frac{{t}}{{t}}\sim\mathrm{1} \\ $$$${conclusion}\:{li}\underset{{x}\rightarrow\mathrm{1}} {{m}}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}+{x}}\right)^{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}−{x}}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by mathdanisur last updated on 29/Jul/21
Thank you Ser, sqr(2/3)
$${Thank}\:{you}\:{Ser},\:{sqr}\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by ArielVyny last updated on 29/Jul/21
yes you have rigth
$${yes}\:{you}\:{have}\:{rigth} \\ $$
Answered by Rasheed.Sindhi last updated on 29/Jul/21
lim_(x→1) (((1+x)/(2+x)))^((1−(√x))/(1−x)) =lim_(x→1) (((1+x)/(2+x)))^((1−(√x))/(1−((√x))^2 )) =lim_(x→1) (((1+x)/(2+x)))^((1−(√x))/((1−(√x))(1+(√x)))) =lim_(x→1) (((1+x)/(2+x)))^(1/(1+(√x))) =(((1+1)/(2+1)))^(1/(1+(√1))) =((2/3))^(1/2) =(√(2/3))
$$\underset{{x}\rightarrow\mathrm{1}} {{lim}}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}+{x}}\right)^{\frac{\mathrm{1}−\sqrt{\boldsymbol{{x}}}}{\mathrm{1}−\boldsymbol{{x}}}} \: \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {{lim}}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}+{x}}\right)^{\frac{\mathrm{1}−\sqrt{\boldsymbol{{x}}}}{\mathrm{1}−\left(\sqrt{{x}}\right)^{\mathrm{2}} }} \: \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {{lim}}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}+{x}}\right)^{\frac{\cancel{\mathrm{1}−\sqrt{\boldsymbol{{x}}}}}{\left(\cancel{\mathrm{1}−\sqrt{{x}}\right)}\left(\mathrm{1}+\sqrt{{x}}\right)}} \: \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {{lim}}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}+{x}}\right)^{\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{x}}}} \:=\left(\frac{\mathrm{1}+\mathrm{1}}{\mathrm{2}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{1}}}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{2}} \\ $$$$=\sqrt{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 29/Jul/21
Thank you Ser
$${Thank}\:{you}\:{Ser} \\ $$
Answered by liberty last updated on 30/Jul/21
T=lim_(x→1) (((2+x−1)/(2+x)))^((1−(√x))/((1+(√x))(1−(√x)))) T= lim_(x→1) (1−(1/(2+x)))^(1/(1+(√x))) T=(1−(1/3))^(1/2) =((2/3))^(1/2) =(√(2/3))=((√6)/3)
$$\:\mathrm{T}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{2}+\mathrm{x}−\mathrm{1}}{\mathrm{2}+\mathrm{x}}\right)^{\frac{\mathrm{1}−\sqrt{\mathrm{x}}}{\left(\mathrm{1}+\sqrt{\mathrm{x}}\right)\left(\mathrm{1}−\sqrt{\mathrm{x}}\right)}} \\ $$$$\mathrm{T}=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}+\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{x}}}} \\ $$$$\:\mathrm{T}=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}=\frac{\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$
Commented by mathdanisur last updated on 30/Jul/21
Thank you Ser
$${Thank}\:{you}\:{Ser} \\ $$
Answered by mathmax by abdo last updated on 31/Jul/21
f(x)=(((x+1)/(x+2)))^((1−(√x))/(1−x)) changement 1−x=t give f(x)=f(1−t)=(((2−t)/(3−t)))^((1−(√(1−t)))/t) ⇒ln(f(1−t)=((1−(√(1−t)))/t)ln(((2−t)/(3−t))) =((1−(√(1−t)))/t){ln(2(1−(t/2)))−ln(3(1−(t/3)))} =((1−(√(1−t)))/t){ln((2/3))+ln(1−(t/2))−ln(1−(t/3))} (t→0) ⇒ln(f(1−t))∼((1−(1−(t/2)))/t){ln((2/3))−(t/2)+(t/3)} ⇒ln(f(1−t))∼(1/2){ln((2/3))−(t/6)}→ln((√(2/3))) ⇒ f(1−t)→(√(2/3)) ⇒lim_(x→1) f(x)=(√(2/3))
$$\mathrm{f}\left(\mathrm{x}\right)=\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}+\mathrm{2}}\right)^{\frac{\mathrm{1}−\sqrt{\mathrm{x}}}{\mathrm{1}−\mathrm{x}}} \:\:\mathrm{changement}\:\mathrm{1}−\mathrm{x}=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{1}−\mathrm{t}\right)=\left(\frac{\mathrm{2}−\mathrm{t}}{\mathrm{3}−\mathrm{t}}\right)^{\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{t}}}{\mathrm{t}}} \Rightarrow\mathrm{ln}\left(\mathrm{f}\left(\mathrm{1}−\mathrm{t}\right)=\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{t}}}{\mathrm{t}}\mathrm{ln}\left(\frac{\mathrm{2}−\mathrm{t}}{\mathrm{3}−\mathrm{t}}\right)\right. \\ $$$$=\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{t}}}{\mathrm{t}}\left\{\mathrm{ln}\left(\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{2}}\right)\right)−\mathrm{ln}\left(\mathrm{3}\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{3}}\right)\right)\right\} \\ $$$$=\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{t}}}{\mathrm{t}}\left\{\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{2}}\right)−\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{3}}\right)\right\}\:\:\left(\mathrm{t}\rightarrow\mathrm{0}\right) \\ $$$$\Rightarrow\mathrm{ln}\left(\mathrm{f}\left(\mathrm{1}−\mathrm{t}\right)\right)\sim\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{2}}\right)}{\mathrm{t}}\left\{\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\frac{\mathrm{t}}{\mathrm{2}}+\frac{\mathrm{t}}{\mathrm{3}}\right\} \\ $$$$\Rightarrow\mathrm{ln}\left(\mathrm{f}\left(\mathrm{1}−\mathrm{t}\right)\right)\sim\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\frac{\mathrm{t}}{\mathrm{6}}\right\}\rightarrow\mathrm{ln}\left(\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{1}−\mathrm{t}\right)\rightarrow\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)=\sqrt{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$
Commented by mathdanisur last updated on 01/Aug/21
Thank You Ser
$${Thank}\:{You}\:{Ser} \\ $$

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