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Question Number 93821 by john santu last updated on 15/May/20

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

Commented by JDamian last updated on 15/May/20

$$0$$

Commented by mathmax by abdo last updated on 15/May/20

$$f\left(x\right)=ln(e^x+1)-x\Rightarrow f\left(x\right)=ln\left(e^x(1+e^{-x})\right)-x$$$$=x+ln(1+e^{-x})-x=ln(1+e^{-x})\Rightarrow {lim}_{x\to +\mathrm{\infty }}f\left(x\right)=0$$

Commented by Kunal12588 last updated on 15/May/20

$$f\left(x\right)=ln(e^x+1)-x=ln(e^x+1)-ln\left(e^x\right)$$$$=ln(1+e^{-x})\Rightarrow \underset{x\to {\mathrm{\infty }}^{+}}{limf}\left(x\right)=0$$

Answered by john santu last updated on 15/May/20

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