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Question Number 202466 by Rydel last updated on 27/Dec/23

$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{x\sqrt{\ln (x^2+1)}}{1+e^{x-3}}$$

Answered by Mathspace last updated on 27/Dec/23

$$={lim}_{x\to +\mathrm{\infty }}{xe}^{-x+3}\sqrt{ln(1+x^2)}/1+e^{-x+3}$$$$={lim}_{x\to +\mathrm{\infty }}{xe}^{-x+3}\sqrt{2lnx+ln(1+\frac{1}{x^2})}$$$$={lim}_{x\to +\mathrm{\infty }}{xe}^{-x+3}\sqrt{2lnx}$$$$=0thebosshereis{e}^{-x}$$

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