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.
Best Answer
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