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Question Number 166975 by henderson last updated on 03/Mar/22

$$\mathrm{hi}!$$$$\mathrm{help}\mathrm{me}!$$$$\underset{\mathit{x}\to -\mathrm{\infty }}{\mathit{l}\mathit{i}\mathit{m}}\frac{{\mathit{e}}^{\frac{1}{\mathit{x}+\sqrt{{\mathit{x}}^2+1}}}}{\mathit{x}}=???$$

Commented by null last updated on 04/Mar/22

$$\mathrm{no}$$

Commented by henderson last updated on 04/Mar/22

$$\mathbf{no}\mathbf{answer}\mathbf{for}\mathbf{this}???:($$

Commented by MJS_new last updated on 04/Mar/22

$$\mathbb{OBVIOUSLY}!!!$$

Answered by null last updated on 04/Mar/22

$$\mathrm{post}1+1\mathrm{instead}!$$$$\mathrm{this}\mathrm{is}\mathrm{wrong}!$$

Answered by MJS_new last updated on 04/Mar/22

$$\underset{x\to -\mathrm{\infty }}{\lim }\frac{{\mathrm{e}}^{\frac{1}{x+\sqrt{x^2+1}}}}{x}=$$$$[t=\frac{1}{x+\sqrt{x^2+1}}\iff x=\frac{1-t^2}{2t};x\to -\mathrm{\infty }\Rightarrow t\to +\mathrm{\infty }]$$$$=\underset{t\to +\mathrm{\infty }}{\lim }\frac{2t{\mathrm{e}}^t}{1-t^2}=\underset{t\to +\mathrm{\infty }}{\lim }\frac{\frac{d^2}{{dt}^2}\left[2t{\mathrm{e}}^t\right]}{\frac{d^2}{{dt}^2}[1-t^2]}=$$$$=\underset{t\to +\mathrm{\infty }}{\lim }-(t+2){\mathrm{e}}^t=-\mathrm{\infty }$$

Commented by MJS_new last updated on 04/Mar/22

$$\mathrm{no}\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{this}???:($$

Commented by henderson last updated on 05/Mar/22

$$\mathrm{thank}\mathrm{you},\mathrm{sir}!$$

Commented by MJS_new last updated on 05/Mar/22

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Answered by Mathspace last updated on 05/Mar/22

$$f\left(x\right)=\frac{e^{\frac{1}{x+\sqrt{x^2+1}}}}{x}{=}_{x=-t}\frac{e^{\frac{1}{-t+\sqrt{t^2+1}}}}{-t}$$$$=-\frac{e^{\frac{1}{\sqrt{t^2+1}-t}}}{t}=-\frac{e^{\sqrt{t^2+1}+t}}{t}(t\to +\mathrm{\infty })$$$$letusehospitaltheorem$$$${lim}_{t\to +\mathrm{\infty }}\frac{e^{\sqrt{t^2+1}+t}}{t}$$$$={lim}_{t\to +\mathrm{\infty }}\frac{d}{dt}\left(e^{\sqrt{t^2+1}+t}\right)$$$$={lim}_{t\to +\mathrm{\infty }}(1+\frac{t}{\sqrt{t^2+t}})e^{\sqrt{t^2+1}+t}$$$$=+\mathrm{\infty }\Rightarrow {lim}_{x\to -\mathrm{\infty }}f\left(x\right)=-\mathrm{\infty }$$

Commented by henderson last updated on 05/Mar/22

$$\mathrm{thank}\mathrm{you},\mathrm{sir}!$$

Commented by henderson last updated on 05/Mar/22

$$\mathrm{i}\mathrm{thought}\mathrm{the}\mathrm{answer}\mathrm{was}\underset{\mathit{x}\to -\mathrm{\infty }}{\mathit{l}\mathit{i}\mathit{m}}\mathit{f}\left(\mathit{x}\right)=0.$$

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