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Question Number 165698 by azizmathhacker last updated on 06/Feb/22

Answered by alephzero last updated on 06/Feb/22

$$\underset{x\to \mathrm{\infty }}{\lim }{e}^{\ln \frac{x+1}{x}}=e^{\ln 1}=1$$

Commented by chrisbridge last updated on 06/Feb/22

$$lim{e}^{\ln \frac{x+1}{x}=lim\frac{x+1}{x}=\frac{x}{x}+\frac{1}{x}}$$$$x\to \mathrm{\infty }$$$$=1+\frac{1}{x}=1+\frac{1}{\mathrm{\infty }}=1+0=1$$$$$$

Commented by alephzero last updated on 07/Feb/22

$$\mathrm{Oh},\mathrm{sorry}.\mathrm{I}\mathrm{forgot}\mathrm{about}\mathrm{that}$$

Commented by TheSupreme last updated on 07/Feb/22

$$e^{ln\left(f\left(x\right)\right)}=f\left(x\right)$$$$lim{e}^{ln\left(f\left(x\right)\right)}=limf\left(x\right)$$$$$$

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