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Question Number 135742 by mohssinee last updated on 15/Mar/21

Commented by mohssinee last updated on 15/Mar/21

$$helpme!$$

Answered by mathmax by abdo last updated on 15/Mar/21

$${\mathrm{e}}^{\frac{1}{\mathrm{x}}}\sim 1+\frac{1}{\mathrm{x}}\mathrm{and}{\mathrm{e}}^{\frac{1}{\mathrm{x}+1}}\sim 1+\frac{1}{\mathrm{x}+1}\Rightarrow {\mathrm{e}}^{\frac{1}{\mathrm{x}}}-{\mathrm{e}}^{\frac{1}{\mathrm{x}+1}}\sim \frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1}=\frac{1}{{\mathrm{x}}^2+\mathrm{x}}\Rightarrow$$$${\mathrm{x}}^3({\mathrm{e}}^{\frac{1}{\mathrm{x}}}-{\mathrm{e}}^{\frac{1}{\mathrm{x}+1}})\sim \frac{{\mathrm{x}}^3}{{\mathrm{x}}^2+\mathrm{x}}\sim \mathrm{x}\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}{\mathrm{x}}^3({\mathrm{e}}^{\frac{1}{\mathrm{x}}}-{\mathrm{e}}^{\frac{1}{\mathrm{x}+1}})=+\mathrm{\infty }$$$$\mathrm{another}\mathrm{method}\mathrm{put}\frac{1}{\mathrm{x}}=\mathrm{t}(\mathrm{sot}\to 0^{+})\Rightarrow \mathrm{x}=\frac{1}{\mathrm{t}}\Rightarrow \mathrm{x}+1=\frac{1}{\mathrm{t}}+1=\frac{\mathrm{t}+1}{\mathrm{t}}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{t}}^3}({\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{t}+1}})\mathrm{we}\mathrm{have}{\mathrm{e}}^{\mathrm{t}}\sim 1+\mathrm{t}+\frac{{\mathrm{t}}^2}{2}$$$${\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{t}+1}}\sim 1+\frac{\mathrm{t}}{\mathrm{t}+1}+\frac{{\mathrm{t}}^2}{2{(\mathrm{t}+1)}^2}\Rightarrow {\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{t}+1}}\sim \mathrm{t}+\frac{{\mathrm{t}}^2}{2}-\frac{\mathrm{t}}{\mathrm{t}+1}-\frac{{\mathrm{t}}^2}{2{(\mathrm{t}+1)}^2}$$$$=\frac{{\mathrm{t}}^2}{\mathrm{t}+1}+\frac{{\mathrm{t}}^2}{2}(1-\frac{1}{{\mathrm{t}}^2+2\mathrm{t}+1})=\frac{{\mathrm{t}}^2}{\mathrm{t}+1}+\frac{{\mathrm{t}}^2}{2}\left(\frac{{\mathrm{t}}^2+2\mathrm{t}}{{\mathrm{t}}^2+2\mathrm{t}+1}\right)\Rightarrow$$$$\frac{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{t}+1}}}{{\mathrm{t}}^3}\sim \frac{1}{\mathrm{t}(\mathrm{t}+1)}+\frac{1}{2\mathrm{t}}\left(\frac{{\mathrm{t}}^2+2\mathrm{t}}{{\mathrm{t}}^2+2\mathrm{t}+1}\right)=\frac{1}{\mathrm{t}(\mathrm{t}+1)}+\frac{\mathrm{t}+2}{2({\mathrm{t}}^2+2\mathrm{t}+1)}$$$${\lim }_{\mathrm{t}\to 0^{+}}\frac{1}{\mathrm{t}(\mathrm{t}+1)}=+\mathrm{\infty }\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=+\mathrm{\infty }$$

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