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I'm trying to evaluate $\displaystyle\lim_{x\to\infty} e^x[e^x\log(1-e^{-x}) - \log(1-e^{-x})+ 1]$.

Using L'hopital's rule I know $\displaystyle\lim_{x\to\infty} e^x\log(1-e^{-x}) = -1$ and so I have a $\infty$ times $0$ situation.

My problem is that when I use L'hopital's rule on the entire expression $e^x[e^x\log(1-e^{-x}) - \log(1-e^{-x})+ 1]$ I seem to be going in an infinite loop.

If I try to evaluate it this way: $\frac{e^x\log(1-e^{-x}) - \log(1-e^{-x})+ 1}{e^{-x}}$ then I always end up with $e^x\log(1-e^{-x})$ on the numerator and $e^{-x}$ on the denominator no matter how many times I take derivatives. These 2 will never simplify to give a nice solution.

If I try the other way and evaluate it this way: $\frac{e^x}{\frac{1}{e^x\log(1-e^{-x}) - \log(1-e^{-x})+ 1}}$ then I always end up with $e^x[e^x\log(1-e^{-x}) - \log(1-e^{-x})+ 1]^n$ for some integer $n$, which is basically the same as where I started.

Are there any other methods that can be used to evaluate this kind of limit? I don't have anything else in my bag of tricks.

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  • $\begingroup$ Aside: this is a limit of the form $\infty \cdot 0$, not a limit of $\infty \cdot 0$. $\endgroup$ – Hurkyl Oct 4 '15 at 11:43
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Hint. You may just put $u=e^{-x}$, getting $$ e^x[e^x\log(1-e^{-x}) - \log(1-e^{-x})+ 1]=\frac1u\left( \frac1u\log(1-u)-\log(1-u)+1\right) $$ then, as $x \to +\infty$, that is $u \to 0^+$, use the Taylor expansion: $$ \log (1-u)=-u-\frac{u^2}2+O(u^3) $$ to get

$$ \lim_{x\to +\infty} e^x[e^x\log(1-e^{-x}) - \log(1-e^{-x})+ 1]= \frac12. $$

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