Question Number 174374 by Best1 last updated on 31/Jul/22
![lim_(x→∞) (((x+3)/(x−2)))^x](https://www.tinkutara.com/question/Q174374.png)
$${li}\underset{{x}\rightarrow\infty} {{m}}\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\right)^{{x}} \\ $$
Answered by LEKOUMA last updated on 31/Jul/22
![lim_(x→∞) [f(x)]^(g(x)) lim_(x→∞) f(x)=1 et lim_(x→∞) g(x)=∞ lim_(x→∞) [f(x)]^(g(x)) =1^∞ (F.I) lim_(x→∞) [f(x)]^(g(x)) =e^(lim_(x→∞) [f(x)−1]g(x)) lim_(x→∞) (((x+3)/(x−2)))^x =1^∞ (F.I) e^(lim_(x→∞) (((x+3)/(x−2))−1)(x)) =e^(lim_(x→∞) (((x+3−1(x−2))/(x−2)))(x)) =e^(lim_(x→∞) (((x+3−x+2)/(x−2)))(x)) =e^(lim_(x→∞) ((5/(x−2)))(x)) =e^(lim_(x→∞) ((5x)/(x−2))) =e^(lim_(x→∞) ((5x)/x)) =e^(lim_(x→∞) 5) =e^5 Donc: lim_(x→∞) (((x+3)/(x−2)))^x =e^5](https://www.tinkutara.com/question/Q174380.png)
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[{f}\left({x}\right)\right]^{{g}\left({x}\right)} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{1}\:{et}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{g}\left({x}\right)=\infty \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[{f}\left({x}\right)\right]^{{g}\left({x}\right)} =\mathrm{1}^{\infty} \left({F}.{I}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[{f}\left({x}\right)\right]^{{g}\left({x}\right)} ={e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[{f}\left({x}\right)−\mathrm{1}\right]{g}\left({x}\right)} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\right)^{{x}} =\mathrm{1}^{\infty} \left({F}.{I}\right) \\ $$$${e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}−\mathrm{1}\right)\left({x}\right)} ={e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}+\mathrm{3}−\mathrm{1}\left({x}−\mathrm{2}\right)}{{x}−\mathrm{2}}\right)\left({x}\right)} \\ $$$$={e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}+\mathrm{3}−{x}+\mathrm{2}}{{x}−\mathrm{2}}\right)\left({x}\right)} ={e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{5}}{{x}−\mathrm{2}}\right)\left({x}\right)} \\ $$$$={e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{5}{x}}{{x}−\mathrm{2}}} ={e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{5}{x}}{{x}}} ={e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}5}} ={e}^{\mathrm{5}} \\ $$$${Donc}:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\right)^{{x}} ={e}^{\mathrm{5}} \\ $$
Commented by Best1 last updated on 31/Jul/22
![lim_(x→∞) (((x+3)/(x−2)))^x A.(1/e) B.1 C.e D.e^5 thank you very much](https://www.tinkutara.com/question/Q174384.png)
$${li}\underset{{x}\rightarrow\infty} {{m}}\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\right)^{{x}} \:\:\:\:{A}.\frac{\mathrm{1}}{{e}}\:\:{B}.\mathrm{1}\:\:{C}.{e}\:\:\:\:{D}.{e}^{\mathrm{5}} \\ $$$${thank}\:{you}\:{very}\:{much}\: \\ $$
Commented by Best1 last updated on 31/Jul/22
![thanks](https://www.tinkutara.com/question/Q174388.png)
$${thanks} \\ $$
Answered by CElcedricjunior last updated on 31/Jul/22
![lim_(x→∞) (((x+3)/(x−2)))^x =1^∞ =FI on a :((x+3)/(x−2))=(((x−2)+5)/(x−2))=1+(5/(x−2)) posons (5/(x−2))=a=>x=(5/a)+2 qd : { ((x−>∞^](https://www.tinkutara.com/question/Q174429.png)