Evaluate the following limits. lim_{(x,y)rightarrow(1,2)}frac{sqrt{y}-sqrt{x+1}}{y-x-1}

We have to find the limit:
$\underset{(x,y)\to (1,2)}{lim}\frac{\sqrt{y}-\sqrt{x+1}}{y-x-1}$
It is $\frac{0}{0}$ form, since there is square root in numerator so applying rationalization rule multiplying and dividing by $\sqrt{y}+\sqrt{x+1}$,
$\underset{(x,y)\to (1,2)}{lim}\frac{\sqrt{y}-\sqrt{x+1}}{(y-x-1)}=\underset{(x,y)\to (1,2)}{lim}\frac{(\sqrt{y}-\sqrt{x+1})}{(y-x-1)}\times \frac{(\sqrt{y}+\sqrt{x+1})}{(\sqrt{y}+\sqrt{x+1})}$
Since we know the identity, $(a-b)(a+b)=a^2-b^2$
Simplifying further,
$\underset{(x,y)\to (1,2)}{lim}\frac{(\sqrt{y}{)}^2-(\sqrt{x+1}{)}^2}{(y-x-1)(\sqrt{y}+\sqrt{x+1})}=\underset{(x,y)\to (1,2)}{lim}\frac{y-(x+1)}{(y-x-1)(\sqrt{y}+\sqrt{x+1})}$
$=\underset{(x,y)\to (1,2)}{lim}\frac{(y-x-1)}{(y-x-1)(\sqrt{y}-\sqrt{x+1})}$
$=\underset{(x,y)\to (1,2)}{lim}\frac{1}{(\sqrt{y}+\sqrt{x+1})}$
$=\frac{1}{(\sqrt{2}+\sqrt{1+1})}$
$=\frac{1}{(\sqrt{2}+\sqrt{2})}$
$=\frac{1}{2\sqrt{2}}$
Hence, value of limits is $\frac{1}{2\sqrt{2}}$