Evaluate \( \lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{x-1} \)

To evaluate the limit \[ \lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{x-1}, \] we start by substituting \( x = 1 \): \[ \frac{\sqrt{1} - 1}{1 - 1} = \frac{0}{0}. \] This indicates that we have an indeterminate form. We can resolve this by using algebraic manipulation. Specifically, we can multiply the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{x} + 1 \): \[ \frac{\sqrt{x}-1}{x-1} \cdot \frac{\sqrt{x}+1}{\sqrt{x}+1} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{(x-1)(\sqrt{x}+1)}. \] The numerator simplifies as follows: \[ (\sqrt{x}-1)(\sqrt{x}+1) = x - 1, \] so we have: \[ \frac{x - 1}{(x-1)(\sqrt{x}+1)} = \frac{1}{\sqrt{x}+1}, \quad \text{for } x \neq 1. \] Now we can evaluate the limit as \( x \) approaches 1: \[ \lim _{x \rightarrow 1} \frac{1}{\sqrt{x}+1} = \frac{1}{\sqrt{1}+1} = \frac{1}{1+1} = \frac{1}{2}. \] Thus, the limit is \[ \boxed{\frac{1}{2}}. \]