Prove that $\lim_{x \to \infty} \frac{\log(1+e^x)}{x} = 1$

Question

Show that $$\lim_{x \to \infty} \frac{\log(1 + e^x)}{x} = 1$$ How do I prove this? Or how do we get this result? Here $\log$ is the natural logarithm.

Answer 1

Hint: How about using $\displaystyle e^x < 1+e^x < 2e^x$, for $x > 0$, taking $\log$ and apply the Squeeze Theorem to conclude the required limit is $1$. (which is essentially same as the other hints).

Answer 2

Hint:

$$\log(1+e^x) = \log(e^x) + (\log(1+e^x) - \log(e^x))=x + \log\left(\frac{1+e^x}{e^x}\right).$$

Answer 3

Hint: Use that $$\log(1 + e^x) = \log[e^x (1 + e^{-x})]$$

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

Answer 4

If $x \to +\infty$, By using L'Hospital's rule:

$$=\lim_{x\to +\infty}\frac{e^x}{1+e^x}=1-\lim_{x\to +\infty}\frac{1}{1+e^x}=1$$

If $x\to -\infty$, It equals 0

So I'm sorry to tell you that this limit does not exist!

Answer 5

You may write, as $x$ tends to $+\infty$, $$ \begin{align} \frac{\log (1+e^x)}{x}&=\frac{\log\left(e^x (1+e^{-x})\right)}{x}\\\\ &=\frac{x+\log(1+e^{-x})}{x}\\\\ &=1+\frac{\log(1+e^{-x})}{x}\\\\ &=1+\frac{e^{-x}}{x}+\mathcal{O}\left(\frac{e^{-2x}}{x}\right)\\\\ \end{align} $$ where we have used $$\log(1+u) =u+\mathcal{O}(u^2)$$ for $u$ near $0$.

Answer 6

For $x\to\infty$

$$\log(1+e^x)\sim\log(e^x)=x$$

Therefore, the limit can be easily simplified

$$\lim_{x\to\infty}\frac{\log(1+e^x)}{x}=\lim_{x\to\infty}\frac{x}{x}=1$$