Evaluate the limit: _x 1^- (x-1)e^(1)/(x-1)

Let $t = \frac{1}{x-1}$.
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, $t = \frac{1}{x-1}$ approaches $-\infty$.
So, the limit variable changes from $x \to 1^-$ to $t \to -\infty$.

From the substitution $t = \frac{1}{x-1}$, we can write $x-1 = \frac{1}{t}$.

This is an indeterminate form of type $(-\infty) \cdot 0$. To evaluate it, we can rewrite the expression as a fraction.
Let $u = -t$. As $t \to -\infty$, $u \to \infty$.
The limit becomes:

This is an indeterminate form of type $\frac{\infty}{\infty}$. We can apply L'Hôpital's Rule.

As $u \to \infty$, $e^u$ approaches $\infty$. Therefore, $\frac{1}{e^u}$ approaches 0.