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Question Number 195453 by Rodier97 last updated on 02/Aug/23

$${\lim }_{x\to +\mathrm{\infty }}xln\left(\frac{e^x+1}{e^x-1}\right)?$$

Commented by Frix last updated on 02/Aug/23

$$0$$

Answered by MM42 last updated on 02/Aug/23

$${lim}_{x\to \mathrm{\infty }}ln{\left(\frac{e^x+1}{e^x-1}\right)}^x★$$$$A={\left(\frac{e^x+1}{e^x-1}\right)}^x\Rightarrow {lim}_{x\to \mathrm{\infty }}A=e^{{lim}_{x\to \mathrm{\infty }}(\frac{e^x+1}{e^x-1}-1)x}$$$$=e^{{lim}_{x\to \mathrm{\infty }}(\frac{e^x+1}{e^x-1}-1)x}=e^{{lim}_{x\to \mathrm{\infty }}\left(\frac{2x}{e^x-1}\right)}=e^0=1$$$$\Rightarrow ★=ln1=0✓$$$$$$

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