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.


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

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

