How to evaluate $\,\lim\limits_{x \to 1}\left(x^n-1\right)\ln^m(1-x)\,$ without L'Hopital?

Question

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!

Answer 1

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

Answer 2

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

• How to prove that exponential grows faster than polynomial?
• Prove $\log(x) \lt x^n$ for all $n \gt 0$

Answer 3

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,

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