Answer
$ - \infty $
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {1^ + }} \ln \left( {\sqrt x - 1} \right) \cr
& {\text{Finding the one sided limit}} \cr
& \cr
& {\text{* }}x \to {1^ + }{\text{ taking }}x = 1.001 \cr
& \approx \ln \left( {\sqrt {1.001} - 1} \right) \cr
& \approx - 7.6 \cr
& \cr
& {\text{* }}x \to {1^ + }{\text{ taking }}x = 1.0001 \cr
& \approx \ln \left( {\sqrt {1.0001} - 1} \right) \cr
& \approx - 9.9 \cr
& \cr
& {\text{* }}x \to {1^ + }{\text{ taking }}x = 1.00001 \cr
& \approx \ln \left( {\sqrt {1.00001} - 1} \right) \cr
& \approx - 12.2 \cr
& \cr
& {\text{* }}x \to {1^ + }{\text{ taking }}x = 1.0000001 \cr
& \approx \ln \left( {\sqrt {1.0000001} - 1} \right) \cr
& \approx - 16.2 \cr
& \cr
& {\text{When }}x \to {1^ + }{\text{ }}\mathop {\lim }\limits_{x \to {1^ + }} \ln \left( {\sqrt x - 1} \right){\text{ tends to }} - \infty \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to {1^ + }} \ln \left( {\sqrt x - 1} \right) = - \infty \cr} $$
