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Question Number 101096 by student work last updated on 30/Jun/20

$$\mathrm{find}\mathrm{the}\mathrm{oblique}\mathrm{asymptote}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\cdot {\mathrm{e}}^{\frac{1}{\mathrm{x}}}$$$$\mathrm{i}\mathrm{need}\mathrm{your}\mathrm{help}$$

Commented by student work last updated on 30/Jun/20

$$\mathrm{i}\mathrm{need}\mathrm{soon}$$

Commented by student work last updated on 30/Jun/20

$$\mathrm{who}\mathrm{can}\mathrm{help}\mathrm{me}?$$

Answered by mathmax by abdo last updated on 30/Jun/20

$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)={\lim }_{\mathrm{x}\to +\mathrm{\infty }}{\mathrm{xe}}^{\frac{1}{\mathrm{x}}}={\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{x}=+\mathrm{\infty }$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{x}}={\lim }_{\mathrm{x}\to +\mathrm{\infty }}{\mathrm{e}}^{\frac{1}{\mathrm{x}}}=1$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)-\mathrm{x}={\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{x}{\mathrm{e}}^{\frac{1}{\mathrm{x}}}-\mathrm{x}={\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{x}({\mathrm{e}}^{\frac{1}{\mathrm{x}}}-1)\mathrm{but}$$$${\mathrm{e}}^{\frac{1}{\mathrm{x}}}=1+\frac{1}{\mathrm{x}}+\mathrm{o}\left(\frac{1}{{\mathrm{x}}^2}\right)\Rightarrow {\mathrm{e}}^{\frac{1}{\mathrm{x}}}-1=\frac{1}{\mathrm{x}}+\mathrm{o}\left(\frac{1}{{\mathrm{x}}^2}\right)\Rightarrow \mathrm{x}({\mathrm{e}}^{\frac{1}{\mathrm{x}}}-1)=1+\mathrm{o}\left(\frac{1}{\mathrm{x}}\right)\Rightarrow$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)-\mathrm{x}=1\Rightarrow \mathrm{the}\mathrm{line}\mathrm{y}=\mathrm{x}+1\mathrm{is}\mathrm{oblique}\mathrm{assymptote}\mathrm{to}{\mathrm{C}}_{\mathrm{f}}\mathrm{at}+\mathrm{\infty }$$$$\mathrm{same}\mathrm{result}\mathrm{at}-\mathrm{\infty }$$$$$$

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