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Question Number 153596 by SANOGO last updated on 08/Sep/21

Commented by SANOGO last updated on 08/Sep/21

$$mercibien$$

Commented by SANOGO last updated on 08/Sep/21

$$mercibien$$

Commented by tabata last updated on 08/Sep/21

$$16)\mathit{A}={\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to \mathrm{\infty }}{\left(\frac{{\mathit{e}}^{\mathit{x}}+1}{\mathit{x}+2}\right)}^{\frac{1}{\mathit{x}+1}}\Rightarrow \mathit{l}\mathit{n}\mathit{A}={\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to \mathrm{\infty }}\frac{\mathit{l}\mathit{n}\left(\frac{{\mathit{e}}^{\mathit{x}}+1}{\mathit{x}+2}\right)}{\mathit{x}+1}$$$$$$$$\mathit{l}\mathit{n}\mathit{A}={\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to \mathrm{\infty }}\frac{(\mathit{x}+2){\mathit{e}}^{\mathit{x}}-({\mathit{e}}^{\mathit{x}}+1)}{{(\mathit{x}+2)}^2}\times \frac{(\mathit{x}+2)}{({\mathit{e}}^{\mathit{x}}+1)}$$$$$$$$\mathit{l}\mathit{n}\mathit{A}={\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to \mathrm{\infty }}({\mathit{e}}^{\mathit{x}}-\frac{1}{\mathit{x}+2})=\mathrm{\infty }$$$$$$$$\Rightarrow \mathit{A}={\mathit{e}}^{\mathrm{\infty }}=\mathrm{\infty }$$$$$$$$⟨\mathit{M}.\mathit{T}⟩$$

Commented by tabata last updated on 08/Sep/21

$$17)\mathit{y}={\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to 0}{\left(\mathit{l}\mathit{n}(1+\mathit{x})\right)}^{\frac{1}{\mathit{l}\mathit{n}\mathit{x}}}\Rightarrow \mathit{l}\mathit{n}\mathit{y}={\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to 0}\frac{\mathit{l}\mathit{n}\left(\mathit{l}\mathit{n}(1+\mathit{x})\right)}{\mathit{l}\mathit{n}\mathit{x}}$$$$$$$$\mathit{l}\mathit{n}\mathit{y}={\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to 0}\left(\frac{1}{\frac{(1+\mathit{x})\mathit{l}\mathit{n}(1+\mathit{x})}{\frac{1}{\mathit{x}}}}\right)\Rightarrow \mathit{l}\mathit{n}\mathit{A}={\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to 0}\left(\frac{\mathit{x}}{(\mathit{x}+1)\mathit{l}\mathit{n}(1+\mathit{x})}\right)$$$$$$$$\mathit{l}\mathit{n}\mathit{A}={\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to 0}(1-\frac{1}{(\mathit{x}+1)\mathit{l}\mathit{n}(1+\mathit{x})})$$$$$$$$\mathit{l}\mathit{n}\mathit{A}=1\Rightarrow \mathit{A}=\mathit{e}$$$$$$$$⟨\mathit{M}.\mathit{T}⟩$$

Commented by tabata last updated on 10/Sep/21

$$\mathit{y}\mathit{o}\mathit{u}\mathit{a}\mathit{r}\mathit{e}\mathit{w}\mathit{e}\mathit{l}\mathit{c}\mathit{o}\mathit{m}\mathit{e}$$

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