Find: $\displaystyle \lim_{x\to -\infty} \dfrac{\ln (1+e^x)}{x}$ (no L'Hospital)

I'm getting a hard time solving this limit. The book shows 1 as the answer, Wolfram Alpha shows 0.

I could solve easily another problem when the denominator was $e^x$ but got stuck on this one.

No L'Hospital rule can be used.

Hints and full answers are appreciated. Sorry if this is a duplicate.

  • 1
    $\begingroup$ Are you sure it's $-\infty$ and not $+\infty$? $\endgroup$ – Elie Louis Dec 28 '17 at 15:45
  • 2
    $\begingroup$ $e^x \to 0$ as $x \to -\infty$, so you get $0$ in the numerator. $\endgroup$ – Daniel Fischer Dec 28 '17 at 15:46
  • $\begingroup$ The answers clearly show that there is a typo in the problem statement as written in the book (is indeed $-\infty$)... or the answer presented by the book is wrong. Not unusual :) $\endgroup$ – bluemaster Dec 28 '17 at 16:04
  • $\begingroup$ @bluemaster If you are ok, you can accept the answer and set as solved. Thanks! $\endgroup$ – gimusi Dec 29 '17 at 20:56

For $x\to +\infty$

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

For $x\to -\infty$

$$\dfrac{\ln (1+e^x)}{x}\to\frac{\ln 1}{-\infty}=0$$


Recall that $\lim\limits_{x \rightarrow x_0} = f(x)g(x) = \lim\limits_{x \rightarrow x_0} f(x) \lim\limits_{x \rightarrow x_0} g(x)$ whenever $\lim\limits_{x \rightarrow x_0} f(x)$ and $\lim\limits_{x \rightarrow x_0} g(x)$ exist.

Since $$\lim\limits_{x\rightarrow -\infty}\ln(1 + e^x) = \ln(1) = 0$$ and $$\lim\limits_{x\rightarrow -\infty}\frac{1}{x} = 0$$

We have $$\lim\limits_{x\rightarrow -\infty}\ln(1 + e^x) \frac{1}{x} = \lim\limits_{x\rightarrow -\infty}\ln(1 + e^x)\lim\limits_{x\rightarrow -\infty}\frac{1}{x} = 0$$

As Gimusi's answer shows, this can be used to rigorously verify that $\lim\limits_{x \rightarrow \infty} \frac{\ln(1+e^x)}{x} = 1$. Note the identity $$\ln(1 + e^x) = \ln(e^{-x}\frac{1 + e^x}{e^{-x}}) = \ln(1+e^{-x}) - \ln (e^{-x})$$

Thus $$\lim\limits_{x\rightarrow \infty}\ln(1 + e^x) \frac{1}{x} = \lim\limits_{x\rightarrow \infty} \left(\ln(1+e^{-x}) - \ln (e^{-x})\right) \frac{1}{x} = \lim\limits_{x\rightarrow -\infty} \left(\ln(1+e^{x}) - \ln (e^{x})\right) \frac{1}{-x} = \lim\limits_{x \rightarrow -\infty}\frac{\ln(e^x)}{x} = 1$$ where the last equality follows because limits distribute over addition (so we apply our previous work).

  • $\begingroup$ I think that an issue that was a little confusing in this case is the final ratio $0/-\infty$ which some might think it is an indeterminate form, and it is not, as clearly showed in this SE post: math.stackexchange.com/q/1430929/460565 $\endgroup$ – bluemaster Dec 28 '17 at 18:21

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