Simplify the denominator of the expression.

The denominator is $(x-1) \div (x-1)$. Since $x \to 1^-$, $x$ is approaching 1 but is not equal to 1. Therefore, $x-1 \neq 0$.
For any $a \neq 0$, $a \div a = 1$. So, $(x-1) \div (x-1) = 1$.
The limit expression simplifies to:

Let $u = \frac{1}{x-1}$.
As $x$ approaches 1 from the left side ($x \to 1^-$), the term $x-1$ approaches 0 from the negative side ($x-1 \to 0^-$).
Therefore, $u = \frac{1}{x-1}$ approaches $\frac{1}{0^-}$, which is $-\infty$.

The limit becomes:

As $u$ approaches $-\infty$, the value of $e^u$ approaches 0.