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Question Number 174374 by Best1 last updated on 31/Jul/22

$$li\underset{x\to \mathrm{\infty }}{m}{\left(\frac{x+3}{x-2}\right)}^x$$

Answered by LEKOUMA last updated on 31/Jul/22

$$\underset{x\to \mathrm{\infty }}{\lim }{\left[f\left(x\right)\right]}^{g\left(x\right)}$$$$\underset{x\to \mathrm{\infty }}{\lim }f\left(x\right)=1et\underset{x\to \mathrm{\infty }}{\lim }g\left(x\right)=\mathrm{\infty }$$$$\underset{x\to \mathrm{\infty }}{\lim }{\left[f\left(x\right)\right]}^{g\left(x\right)}=1^{\mathrm{\infty }}(F.I)$$$$\underset{x\to \mathrm{\infty }}{\lim }{\left[f\left(x\right)\right]}^{g\left(x\right)}=e^{\underset{x\to \mathrm{\infty }}{\lim }[f\left(x\right)-1]g\left(x\right)}$$$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{x+3}{x-2}\right)}^x=1^{\mathrm{\infty }}(F.I)$$$$e^{\underset{x\to \mathrm{\infty }}{\lim }(\frac{x+3}{x-2}-1)\left(x\right)}=e^{\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{x+3-1(x-2)}{x-2}\right)\left(x\right)}$$$$=e^{\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{x+3-x+2}{x-2}\right)\left(x\right)}=e^{\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{5}{x-2}\right)\left(x\right)}$$$$=e^{\underset{x\to \mathrm{\infty }}{\lim }\frac{5x}{x-2}}=e^{\underset{x\to \mathrm{\infty }}{\lim }\frac{5x}{x}}=e^{\underset{x\to \mathrm{\infty }}{\lim 5}}=e^5$$$$Donc:\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{x+3}{x-2}\right)}^x=e^5$$

Commented by Best1 last updated on 31/Jul/22

$$li\underset{x\to \mathrm{\infty }}{m}{\left(\frac{x+3}{x-2}\right)}^{x}A.\frac{1}{e}B.1C.eD.e^5$$$$thankyouverymuch$$

Commented by Best1 last updated on 31/Jul/22

$$thanks$$

Answered by CElcedricjunior last updated on 31/Jul/22

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{x+3}{x-2}\right)}^x=1^{\mathrm{\infty }}=\mathit{F}\mathit{I}$$$$ona:\frac{x+3}{x-2}=\frac{(x-2)+5}{x-2}=1+\frac{5}{\mathit{x}-2}$$$$\mathit{p}\mathit{o}\mathit{s}\mathit{o}\mathit{n}\mathit{s}\frac{5}{\mathit{x}-2}=\mathit{a}=>\mathit{x}=\frac{5}{\mathit{a}}+2$$$$\mathit{q}\mathit{d}:\{\begin{array}{l}\mathit{x}->\stackrel{`}{\mathrm{\infty }}\\ \mathit{a}->0\end{array}$$$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{\mathit{x}+3}{\mathit{x}-2}\right)}^{\mathit{x}}=\underset{\mathit{a}\to 0}{\lim }{(1+\mathit{a})}^{\frac{5}{\mathit{a}}+2}$$$$=\underset{\mathit{a}\to 0}{\lim }{\mathit{e}}^{\frac{5\mathit{l}\mathit{n}(1+\mathit{a})}{\mathit{a}}+2\mathit{l}\mathit{n}(1+\mathit{a})}={\mathit{e}}^5$$$$\mathit{c}\mathit{a}\mathit{s}\underset{x\to 0}{\lim }\frac{\mathbf{ln}(1+\mathbf{x})}{\mathbf{x}}=\underset{x\to 0}{\lim }\frac{\mathit{l}\mathit{n}(1+\mathit{x})-\mathit{l}\mathit{n}\left(1\right)}{\mathit{x}-0}={\left(\mathit{l}\mathit{n}(1+\mathit{x})\right)}^{\prime }\left(0\right)=\frac{1}{1+\left(0\right)}=1$$$$$$$$$$$$.....lec\acute{el}\stackrel{`}{ebre}cedricjunior.......99$$

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