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$$\lim_{x\to-\infty} \frac{\ln(e^x+1)}{x}$$

due to Squeeze theorem and $\lim_{x\to0}\frac{\ln(1+x)}{x}=1$ we have

$$1\leftarrow\frac{\ln(e^x+1)}{e^x}\leq \frac{\ln(e^x+1)}{x}$$

what can be the upper bound of this function, or is there better solution?

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  • $\begingroup$ Are you sure about $-\infty$ ? $\endgroup$ – Yves Daoust Jan 11 '17 at 17:44
  • $\begingroup$ I should have the inequality reversely..... $\endgroup$ – UfmdFkiF Jan 12 '17 at 0:02
  • $\begingroup$ Are you sure the problem is asked with $-\infty$ ? This is unlikely. $\endgroup$ – Yves Daoust Jan 12 '17 at 7:56
  • $\begingroup$ You are right @YvesDaoust. shame on me... $\endgroup$ – UfmdFkiF Jan 12 '17 at 18:36
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Consider that $$ \lim\limits_{x \to -\infty}\ln(e^x+1) = \ln\left(\lim\limits_{x \to -\infty} e^x + 1\right)= \ln(1) = 0$$ and $$ \lim\limits_{x \to -\infty}\frac{1}{x}= 0$$ So your limit is $0$.

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$$ \ln(e^x+1)=x+\ln(1+e^{-x}). $$

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You don't even need the Squeeze Theorem here. As $x\to-\infty$, $e^x\to0$, so the numerator in the limit turns into $\ln(0+1)=0$. Therefore, $$\lim_{x\to-\infty} \frac{\ln(e^x+1)}{x}=\frac{0}{-\infty}=0.$$

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Let $x\to -\infty$. Since $\ln(e^x+1)\to\ln 1=0$ and $\frac{1}{x}\to 0$, the limit in question equals to zero.

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