Answer
$1$
Work Step by Step
Here, we have $\lim\limits_{a \to 0^{+}}\int_a^{1} (-\ln x) dx=\lim\limits_{a \to 0^{+}}[x-x\ln x]_a^1$
This implies that
$\lim\limits_{a \to 0^{+}}[x-x\ln x]_a^1=1-\lim\limits_{a \to 0^{+}}a+
\lim\limits_{a \to 0^{+}} a \ln a$
or, $1-\lim\limits_{a \to 0^{+}}a+
\lim\limits_{a \to 0^{+}} a \ln a=1-\lim\limits_{a \to 0^{+}} \ln a$
But $\lim\limits_{a \to 0^{+}} \ln a=-\infty/\infty$
This shows an indeterminate form of the limit, so we will use L-Hospital's rule.
$1-\lim\limits_{a \to 0^{+}} \ln a=1-\lim\limits_{a \to 0^{+}} (-a)$
Thus,
$1-\lim\limits_{a \to 0^{+}} (-a)=1-0=1$
