We are asked to solve the inequality $\frac{|x-1|}{x} < 1$.

1. Problem Description

We are asked to solve the inequality $\frac{|x-1|}{x} < 1$.

2. Solution Steps

First, we must have $x \ne 0$.

Case 1: $x > 0$.

Since $x > 0$, we can multiply both sides of the inequality by $x$ without changing the inequality sign:

$|x-1| < x$

This inequality means that $-x < x-1 < x$. We can split this into two inequalities:

$x-1 < x$ and $-x < x-1$.

The first inequality, $x-1 < x$, simplifies to $-1 < 0$, which is always true.

The second inequality, $-x < x-1$, simplifies to $1 < 2x$, so $x > \frac{1}{2}$.

Thus, in this case, we have $x > \frac{1}{2}$.

Case 2: $x < 0$.

Since $x < 0$, we can multiply both sides of the inequality by $x$, but we must reverse the inequality sign:

$|x-1| > x$

Since $x < 0$, we have $x-1 < -1 < 0$, so $x-1 < 0$.

Therefore, $|x-1| = -(x-1) = 1-x$.

The inequality becomes $1-x > x$.

This simplifies to $1 > 2x$, so $x < \frac{1}{2}$.

Since we are considering $x < 0$, the solution in this case is $x < 0$.

Combining both cases:

From Case 1, we have $x > \frac{1}{2}$.

From Case 2, we have $x < 0$.

Thus, the solution is $x < 0$ or $x > \frac{1}{2}$.

3. Final Answer

$x < 0$ or $x > \frac{1}{2}$

In interval notation, the solution is $(-\infty, 0) \cup (\frac{1}{2}, \infty)$.