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This question arose as part of the evaluation of $\int_{0}^{1} x^{n-1} \ln^{m}(1-x) \, \mathrm{d}x$ using integration by parts, where we would require to show that $$ L=\lim_{x \to 1} \left( x^n-1\right)\ln^m(1-x) \qquad n,m \in \mathbb{N} $$ Checking several cases with Wolfram Alpha, the claim seemed to hold.

My attempt: \begin{align} L & \overset{\color{blue}{x = 1-e^{-t}}}{=}\lim_{t \to \infty} \left(\left(1-e^{-t} \right)^n-1 \right)\left(-t \right)^m\\ & = \sum_{k=1}^{n}\binom{n}{k}(-1)^{k+m}\lim_{t \to \infty}\frac{t^m}{e^{kt}}\\ & \overset{\color{green}{\text{L'H}}}{=} \sum_{k=1}^{n}\binom{n}{k}(-1)^{k+m}\lim_{t \to \infty}\frac{m!}{k^me^{kt}}\\ & =0 \end{align}

I was wondering if the limit in the title could be shown by a method (which could be completely different than the one I tried) that doesn't use L'Hopitals rule. The reason is that I believe another proof with L'H would be similar-ish in spirit to mine, and I was more interested in other ways of tackling the original limit altogether.

Another idea I tried to make work expanding the natural log part in a series, but since the limit is on the border of the convergence region I wasn't sure if it would work.

Any ideas or suggestions are welcome. Thanks!

Robert Lee
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    Can you use the fact that, $\lim\limits_{u\to 0} u \ln^m(u) = 0$? – Kroki Jun 29 '23 at 06:03
  • @Youem, the only way I know to prove your statement is again using L'H. Any alternative proofs are very welcome, but I feel that unless there's an L'H-less proof of your fact it feels a bit like hiding L'H "under the hood". – Robert Lee Jun 29 '23 at 06:10
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    There exist proofs of $\lim_{t \to \infty} \frac{t^m}{e^t} = 0$ which do not use L'Hospital. The best of these that I can think of is this one. – Sarvesh Ravichandran Iyer Jun 29 '23 at 06:19
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    $e^{t} \geq \frac {t^{m+1}} {(m+1)!}$ gives $\lim_{t \to\infty} \frac {t^{m}} {e^{t}}=0$. – geetha290krm Jun 29 '23 at 06:30
  • @SarveshRavichandranIyer, that's a really nice answer. Thanks for the link! – Robert Lee Jun 29 '23 at 06:58
  • @RobertLee You're welcome! – Sarvesh Ravichandran Iyer Jun 29 '23 at 06:58
  • Does this answer your question? What is the workings to solve $\lim\limits_{n\rightarrow \infty} \frac{n^t}{e^n}$? or this? https://math.stackexchange.com/a/2974234 or that? https://math.stackexchange.com/a/3907839 – Anne Bauval Jun 29 '23 at 07:26
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    I was looking at the title & then at "was wondering if the limit could be shown without" , which I assumed to refer to the title. I now see that the title is inaccurate , OP wants the Inner Part of the limit in the Summation. Basically "my attempt" by OP refers to overall title limit & OP has made no attempt on the Inner Part of the limit , @AnneBauval , which is a Dup of other Questions. My Answer has no value here , hence I deleted it ! [ I was aware of various Proofs of the "Polynomial versus Exponential" limit , including with $\log$ & with LH , hence I brushed it off as well known ! ] – Prem Jun 29 '23 at 08:45
  • @Prem, I think my writing of the question was poor and led to a misunderstanding as to what I was looking for. I am only interested in the title limit, not the inner part. My attempt was only included to abide by the site's "show your work" rule, but any other proof of the title, by any other method, that may or may not reduce to the inner-limit I use in my attempt _is_what I was looking for. I also previously found the posts of the limit you link, but I was interested in possible alternative ways of solving the title limit with or without it. I'll edit my question to clarify. Thanks! – Robert Lee Jun 29 '23 at 20:01

3 Answers3

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A simpler solution can be obtained by simple substitutions.

$$(1-x^n)\log^m(1-x)=(1+x+\ldots+x^{n-1})\Big(\sqrt[m]{1-x}\log(1-x)\Big)^m$$

Let $h=1-x$. Then $h\rightarrow0+$ if and only if $x\rightarrow1-$. It follows that $$\begin{align}\ \sqrt[m]{1-x}\log(1-x)=h^{1/m}\log h\xrightarrow{h\rightarrow0+}0\tag{1}\label{limit-log}\end{align}$$ This follows from the well known fact that $\lim_{t\rightarrow\infty}\frac{t^p}{e^{at}}=0$ for all real $p$ and $a>0$ (substitute $t=-\log h$ in $\eqref{limit-log}$).

Therefore $$\lim_{ x\rightarrow1}(1-x^n)\log^m(1-x)=0$$

Mittens
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We have that by geometric series

$$\left( x^n-1\right)=-(1-x)\sum_{k=0}^{n-1}x^k$$

with $-\sum_{k=0}^{n-1}x^k\to -n$, then we reduce to the standard limit

$$\lim_{x \to 1}\;(1-x)\ln^m(1-x)=0$$

which can be proved in several ways, as for example discussed here

user
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Given positive real $m$ and $k$, define $f: \mathbb{R}^+ \to \mathbb{R}$ with

$$ f(t) = t^m e^{-kt}. $$

Also let $g: \mathbb{R}^+ \to \mathbb{R}$ be the logarithm of $f$:

$$ g(t) = \ln f(t) = m\ln t - kt $$

$$ g'(t) = \frac{m}{t} - k $$

Notice when $t > \frac{2m}{k}$, $g'(t) < -\frac{k}{2}$, so

$$ g(t) < \int_{2m/k}^t -\frac{k}{2}\, ds = -\frac{kt}{2} + m $$

This gives $\lim_{t \to +\infty} g(t) = -\infty$. And finally,

$$\lim_{t \to +\infty} f(t) = \lim_{t \to +\infty} \exp(g(t)) = 0. $$

aschepler
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