Menu Close

lim-x-x-2-4-x-2-2-x-2-1-x-1-pls-

Tinku Tara 0 Comments
Pin




Question Number 59393 by Mikael_Marshall last updated on 09/May/19
lim_(x→+∞) (((x^2 −4)/(x^2 +2)))^((x^2 −1)/(x+1)) pls.
$$\underset{{x}\rightarrow+\infty} {{lim}}\:\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{2}}\overset{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}} {\right)} \\ $$$${pls}. \\ $$
Commented by maxmathsup by imad last updated on 10/May/19
let A(x) =(((x^2 −4)/(x^2 +2)))^((x^2 −1)/(x+1)) ⇒ln(A(x)) =(((x^2 −1)/(x^2 +2)))ln(((x^2 −4)/(x^2 +2))) but ln(((x^2 −4)/(x^2 +2))) =ln(((x^2 +2−6)/(x^2 +2)))=ln(1−(6/(x^2 +2)))∼((−6)/(x^2 +2))(x→+∞) (because ln(1+u)∼u (u→0) ⇒ln(A(x))∼((−6(x^2 −1))/((x^2 +2)^2 )) →0 (x→+∞) ⇒lim_(x→+∞) A(x) =1 .
$${let}\:{A}\left({x}\right)\:=\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} \:+\mathrm{2}}\right)^{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}} \:\Rightarrow{ln}\left({A}\left({x}\right)\right)\:=\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{2}}\right){ln}\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{2}}\right)\:\:{but} \\ $$$${ln}\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} \:+\mathrm{2}}\right)\:={ln}\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}−\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{2}}\right)={ln}\left(\mathrm{1}−\frac{\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{2}}\right)\sim\frac{−\mathrm{6}}{{x}^{\mathrm{2}} \:+\mathrm{2}}\left({x}\rightarrow+\infty\right)\:\:\left({because}\:{ln}\left(\mathrm{1}+{u}\right)\sim{u}\:\left({u}\rightarrow\mathrm{0}\right)\right. \\ $$$$\Rightarrow{ln}\left({A}\left({x}\right)\right)\sim\frac{−\mathrm{6}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{2}} }\:\rightarrow\mathrm{0}\:\:\left({x}\rightarrow+\infty\right)\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {A}\left({x}\right)\:=\mathrm{1}\:. \\ $$
Commented by Mikael_Marshall last updated on 10/May/19
thank you Sir
$${thank}\:{you}\:{Sir} \\ $$
Commented by mr W last updated on 10/May/19
please check sir: let A(x) =(((x^2 −4)/(x^2 +2)))^((x^2 −1)/(x+1)) ⇒ln(A(x)) =(((x^2 −1)/(x +1)))ln(((x^2 −4)/(x^2 +2)))
$${please}\:{check}\:{sir}: \\ $$$${let}\:{A}\left({x}\right)\:=\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} \:+\mathrm{2}}\right)^{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}} \:\Rightarrow{ln}\left({A}\left({x}\right)\right)\:=\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}\:+\mathrm{1}}\right){ln}\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{2}}\right)\: \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 10/May/19
if limf(x)^(g(x)) =1^∞ we can use lim e^(g(x)(f(x)−1)) so lim_(x→+∞) e^((((x^2 −1)/(x+1)))(((x^2 −4)/(x^2 +2))−1)) =lim_(x→+∞) e^((((x^2 −1)/(x+1)))(((−6)/(x^2 +2)))) = e^0 =1
$${if}\:{limf}\left({x}\right)^{{g}\left({x}\right)} =\mathrm{1}^{\infty} \:{we}\:{can}\:{use} \\ $$$${lim}\:{e}^{{g}\left({x}\right)\left({f}\left({x}\right)−\mathrm{1}\right)} \\ $$$${so} \\ $$$${lim}_{{x}\rightarrow+\infty} \:{e}^{\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}\right)\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{2}}−\mathrm{1}\right)} ={lim}_{{x}\rightarrow+\infty} \:{e}^{\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}\right)\left(\frac{−\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{2}}\right)} = \\ $$$${e}^{\mathrm{0}} =\mathrm{1} \\ $$
Commented by Mikael_Marshall last updated on 10/May/19
thanks Sir
$${thanks}\:{Sir} \\ $$
Commented by maxmathsup by imad last updated on 10/May/19
yes sir i have commited a error let rectify ln(A(x))=((x^2 −1)/(x+1))ln(((x^2 −4)/(x^2 +2))) ⇒ ln(A(x)) ∼((x^2 −1)/(x+1)).((−6)/(x^2 +2)) =((−6x^2 +6)/((x+1)(x^2 +2))) →0(x→+∞) ⇒lim_(x→+∞) A(x)=1.
$${yes}\:{sir}\:{i}\:{have}\:{commited}\:{a}\:{error}\:{let}\:{rectify}\:\:{ln}\left({A}\left({x}\right)\right)=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}{ln}\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} \:+\mathrm{2}}\right)\:\Rightarrow \\ $$$${ln}\left({A}\left({x}\right)\right)\:\sim\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}.\frac{−\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{2}}\:\:=\frac{−\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)}\:\rightarrow\mathrm{0}\left({x}\rightarrow+\infty\right)\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {A}\left({x}\right)=\mathrm{1}. \\ $$$$ \\ $$
Answered by malwaan last updated on 10/May/19
lim_(x→∞) (((x^2 +2−6)/(x^2 +2)))^(((x+1)(x−1))/(x+1)) =lim_(x→∞) (1−(6/(x^2 +2)))^(x−1) =lim_(x→∞) (1+((−6)/(x^2 +2)))^((((x^2 +2)/(−6)))(x−1)(((−6)/(x^2 +2_ )))) =lim_(x→∞) [(1+((−6)/(x^2 +2)))^((((x^2 +2)/(−6)))) ]^((−6(x−1))/(x^2 +2)) =[e^(−6) ]^0 =1
$$\underset{{x}\rightarrow\infty} {{lim}}\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}−\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{2}}\right)^{\frac{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)}{{x}+\mathrm{1}}} \\ $$$$=\underset{{x}\rightarrow\infty} {{lim}}\left(\mathrm{1}−\frac{\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{2}}\right)^{{x}−\mathrm{1}} \\ $$$$=\underset{{x}\rightarrow\infty} {{lim}}\left(\mathrm{1}+\frac{−\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{2}}\right)^{\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}}{−\mathrm{6}}\right)\left({x}−\mathrm{1}\right)\left(\frac{−\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{2}_{} }\right)} \\ $$$$=\underset{{x}\rightarrow\infty} {{lim}}\left[\left(\mathrm{1}+\frac{−\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{2}}\right)^{\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}}{−\mathrm{6}}\right)} \right]^{\frac{−\mathrm{6}\left({x}−\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{2}}} \\ $$$$=\left[{e}^{−\mathrm{6}} \right]^{\mathrm{0}} \:=\mathrm{1} \\ $$
Commented by Mikael_Marshall last updated on 10/May/19
thanks Sir
$${thanks}\:{Sir} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *