Evaluate \( \lim _{x \rightarrow 1} \frac{x-1}{\sqrt{x}-1} \). If the corresponding limit is infinity, then insert INF in the blank. If the corresponding limit is negative infinity, then insert -INF in the blank. If the corresponding limit does not exist, then insert DNE in the blank. Otherwise, write the number as an integer or a reduced fraction.

To evaluate the limit \[ \lim_{x \rightarrow 1} \frac{x-1}{\sqrt{x}-1}, \] we start by substituting \( x = 1 \). This gives us \[ \frac{1 - 1}{\sqrt{1} - 1} = \frac{0}{0}, \] which is an indeterminate form. Thus, we need to simplify the expression. We can use the identity \( a^2 - b^2 = (a-b)(a+b) \) to simplify the denominator. We rewrite \(\sqrt{x} - 1\) as follows: \[ \sqrt{x} - 1 = \frac{x - 1}{\sqrt{x} + 1}. \] Substituting this back into the limit gives: \[ \lim_{x \rightarrow 1} \frac{x-1}{\sqrt{x} - 1} = \lim_{x \rightarrow 1} \frac{x-1}{\frac{x-1}{\sqrt{x} + 1}} = \lim_{x \rightarrow 1} (x-1) \cdot \frac{\sqrt{x} + 1}{x-1}. \] Now, we can cancel \(x-1\) (noting that \(x \neq 1\) in the limit process): \[ \lim_{x \rightarrow 1} \sqrt{x} + 1. \] Now we find this limit: \[ \sqrt{1} + 1 = 1 + 1 = 2. \] Thus, the limit is: \[ \lim_{x \rightarrow 1} \frac{x-1}{\sqrt{x} - 1} = 2. \] Hence, we insert \( \boxed{2} \) in the blank.