Question Number 174374 by Best1 last updated on 31/Jul/22
$${li}\underset{{x}\rightarrow\infty} {{m}}\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\right)^{{x}} \\ $$
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
$${li}\underset{{x}\rightarrow\infty} {{m}}\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\right)^{{x}} \:\:\:\:{A}.\frac{\mathrm{1}}{{e}}\:\:{B}.\mathrm{1}\:\:{C}.{e}\:\:\:\:{D}.{e}^{\mathrm{5}} \\ $$$${thank}\:{you}\:{very}\:{much}\: \\ $$
Commented by Best1 last updated on 31/Jul/22
$${thanks} \\ $$
Answered by CElcedricjunior last updated on 31/Jul/22
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\right)^{{x}} =\mathrm{1}^{\infty} =\boldsymbol{{FI}} \\ $$$${on}\:{a}\::\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}=\frac{\left({x}−\mathrm{2}\right)+\mathrm{5}}{{x}−\mathrm{2}}=\mathrm{1}+\frac{\mathrm{5}}{\boldsymbol{{x}}−\mathrm{2}} \\ $$$$\boldsymbol{{posons}}\:\frac{\mathrm{5}}{\boldsymbol{{x}}−\mathrm{2}}=\boldsymbol{{a}}=>\boldsymbol{{x}}=\frac{\mathrm{5}}{\boldsymbol{{a}}}+\mathrm{2} \\ $$$$\boldsymbol{{qd}}\::\begin{cases}{\boldsymbol{{x}}−>\grave {\infty}}\\{\boldsymbol{{a}}−>\mathrm{0}}\end{cases} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\boldsymbol{{x}}+\mathrm{3}}{\boldsymbol{{x}}−\mathrm{2}}\right)^{\boldsymbol{{x}}} =\underset{\boldsymbol{{a}}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\boldsymbol{{a}}\right)^{\frac{\mathrm{5}}{\boldsymbol{{a}}}+\mathrm{2}} \\ $$$$=\underset{\boldsymbol{{a}}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{e}}^{\frac{\mathrm{5}\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{a}}\right)}{\boldsymbol{{a}}}+\mathrm{2}\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{a}}\right)} =\boldsymbol{{e}}^{\mathrm{5}} \\ $$$$\boldsymbol{{cas}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{x}}\right)−\boldsymbol{{ln}}\left(\mathrm{1}\right)}{\boldsymbol{{x}}−\mathrm{0}}=\left(\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{x}}\right)\right)'\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{0}\right)}=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$…..{le}\:\:{c}\acute {{e}l}\grave {{e}bre}\:{cedric}\:{junior}…….\mathrm{99} \\ $$