Answer
$$\frac{1}{{2\sqrt 2 }}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{\sqrt y - \sqrt {x + 1} }}{{y - x - 1}} \cr
& {\text{Evaluating the limit}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{\sqrt y - \sqrt {x + 1} }}{{y - x - 1}} = \frac{{\sqrt 2 - \sqrt {1 + 1} }}{{2 - 1 - 1}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{\sqrt y - \sqrt {x + 1} }}{{y - x - 1}} = \frac{0}{0}{\text{Ind}} \cr
& {\text{Rationalizing the numerator}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{\sqrt y - \sqrt {x + 1} }}{{y - x - 1}} \cdot \frac{{\sqrt y + \sqrt {x + 1} }}{{\sqrt y + \sqrt {x + 1} }} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{{{\left( {\sqrt y } \right)}^2} - {{\left( {\sqrt {x + 1} } \right)}^2}}}{{\left( {y - x - 1} \right)\left( {\sqrt y + \sqrt {x + 1} } \right)}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{y - x - 1}}{{\left( {y - x - 1} \right)\left( {\sqrt y + \sqrt {x + 1} } \right)}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{1}{{\sqrt y + \sqrt {x + 1} }} \cr
& {\text{Evaluating}} \cr
& = \frac{1}{{\sqrt 2 + \sqrt {1 + 1} }} \cr
& = \frac{1}{{2\sqrt 2 }} \cr} $$
