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

To solve the limit \( \lim _{x \rightarrow 1} \frac{x-1}{\sqrt{x}-1} \), we can start by substituting \( x = 1 \): \[ \frac{1-1}{\sqrt{1}-1} = \frac{0}{0} \] This is an indeterminate form, so we need to manipulate the expression further. We can multiply the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{x} + 1 \): \[ \lim _{x \rightarrow 1} \frac{x-1}{\sqrt{x}-1} \cdot \frac{\sqrt{x}+1}{\sqrt{x}+1} = \lim _{x \rightarrow 1} \frac{(x-1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)} \] Notice that \( (\sqrt{x}-1)(\sqrt{x}+1) = x - 1 \). So we can rewrite the limit as follows: \[ \lim _{x \rightarrow 1} \frac{(x-1)(\sqrt{x}+1)}{x-1} \] Provided \( x \neq 1 \), we can cancel \( x-1 \) in the numerator and denominator: \[ \lim _{x \rightarrow 1} (\sqrt{x}+1) \] Now, we can substitute \( x = 1 \): \[ \sqrt{1}+1 = 1 + 1 = 2 \] Thus, the limit is: \[ \lim _{x \rightarrow 1} \frac{x-1}{\sqrt{x}-1} = 2 \]