Integral limit without using Taylor expansion I’m trying to compute
$$\lim_{y \to 1-} (1-y + \ln (y))\int_0^y \frac{dx}{(x-1) \ln(x)}$$
I was able to show with Taylor series that this converges to 0, but it was tedious .
Is there a more elegant way to do this perhaps using upper and lower bounds?
Thank you.
 A: 
Herein, we present a solution that relies only on elementary inequalities and the squeeze theorem.  To that end we now proceed.


Let $I(y)$ be the integral given by
$$I(y)=\int_0^y \frac{1}{(x-1)\log(x)}\,dx\tag 1$$
for $y\in (0,1)$.


ASIDE:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1}\tag 2$$


Rearranging the inequalities in $(2)$, we see that for $x<1$
$$\frac{1}{x-1}\le \frac{1}{\log(x)}\le \frac{x}{x-1} \tag3$$

Using $(3)$ in $(1)$, we find that
$$\frac{y}{1-y} +\log(1-y)=\int_0^y \frac{x}{(x-1)^2}\,dx\le I(y)\le \int_0^y \frac{1}{(x-1)^2}\,dx =\frac{y}{1-y} \tag 4$$

Multiplying $(4)$ by $(1-y+\log(y))$ reveals
$$(1-y+\log(y))\left(\frac{y}{1-y} +\log(1-y)\right)\le (1-y+\log(y))I(y)\le (1-y+\log(y))\left(\frac{y}{1-y}\right)\tag 5$$
whereupon applying the squeeze theorem to $(5)$ yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 1^-}(1-y+\log(y))\int_0^y \frac{1}{(x-1)\log(x)}\,dx=0}$$

A: Substitute $y=1-t$, so the limit becomes
$$
\lim_{t\to0^+}(t+\ln(1-t))\int_0^{1-t}\frac{1}{(x-1)\ln x}\,dx=
\lim_{t\to0^+}-\frac{t^2}{2}\int_0^{1-t}\frac{1}{(x-1)\ln x}\,dx
$$
because
$$
\lim_{t\to0^+}\frac{t+\ln(1-t)}{t^2}=-\frac{1}{2}
$$
In the integral, do the substitution $x=1-u$, so we get
$$
\lim_{t\to0^+}\frac{t^2}{2}\int_t^1\frac{1}{u\ln(1-u)}\,du
$$
Now change $t=1/s$, so the limit is
$$
\lim_{s\to\infty}
\frac{\displaystyle\int_{1/s}^1\dfrac{1}{u\ln(1-u)}\,du}{2s^2}=
\lim_{s\to\infty}\dfrac{\dfrac{1}{s^2}\dfrac{1}{(1/s)\ln(1-1/s)}}{4s}=
\lim_{t\to0^+}\frac{t^2}{4\ln(1-t)}=0
$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
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$\ds{\lim_{y \to 1^{\large-}}\braces{\bracks{1 - y + \ln\pars{y}}
\int_{0}^{y}{\dd x \over \pars{x - 1}\ln\pars{x}}}:\ {\large ?}}$.

\begin{align}
&\int_{0}^{y}{\dd x \over \pars{x - 1}\ln\pars{x}} =
\int_{0}^{y}\bracks{{1 \over \pars{x - 1}\ln\pars{x}} -
{1 \over \pars{x - 1}^{2}} - {1 \over 2\pars{x - 1}}}\dd x
\\[2mm] + &\
\int_{0}^{y}\bracks{
{1 \over \pars{x - 1}^{2}} + {1 \over 2\pars{x - 1}}}\dd x
\\[1cm] = &\
\int_{0}^{y}\bracks{{1 \over \pars{x - 1}\ln\pars{x}} -
{1 \over \pars{x - 1}^{2}} - {1 \over 2\pars{x - 1}}}\dd x
\\[2mm] - &\ {1 \over y - 1} - 1 + {1 \over 2}\ln\pars{1 - y} \sim
c - {1 \over y - 1} + {1 \over 2}\ln\pars{1 - y}\quad
\mbox{as}\ y \to 1^{\large -}\,,\ \pars{~c:\ constant~}
\end{align}

and
  $\ds{1 - y + \ln\pars{y} \sim -\,{1 \over 2}\pars{y - 1}^{2}
\quad\mbox{as}\ y \to 1^{\large -}}$.


Then,
$$
\bbx{\lim_{y \to 1^{\large-}}\braces{\bracks{1 - y + \ln\pars{y}}
\int_{0}^{y}{\dd x \over \pars{x - 1}\ln\pars{x}}} = 0}
$$
