Given:

\(\lim_{(x,y)\rightarrow(3,3)}(\frac{x-y}{\sqrt x-\sqrt y})\)

On putting the limits directly, we get:

\(\lim_{(x,y)\rightarrow(3,3)}(\frac{x-y}{\sqrt x-\sqrt y})=\frac{0}{0}\)

On further solving the equation, we have:

\(\lim_{(x,y)\rightarrow(3,3)}(\frac{x-y}{\sqrt x-\sqrt y})=\lim_{(x,y)\rightarrow(3,3)}(\frac{(\sqrt x)^2-(\sqrt y)^2}{\sqrt x-\sqrt y})\)

Since,

\(a^2-b^2=(a-b)(a+b)\)

Therefore,

\(\lim_{(x,y)\rightarrow(3,3)}(\frac{(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)}{\sqrt x-\sqrt y})\)

\(\lim_{(x,y)\rightarrow(3,3)}(\sqrt x+\sqrt y)\)

Now putting the limits, we get:

\((\sqrt3+\sqrt3)\)

\(=2\sqrt3\)

The required answer.

\(\lim_{(x,y)\rightarrow(3,3)}(\frac{x-y}{\sqrt x-\sqrt y})\)

On putting the limits directly, we get:

\(\lim_{(x,y)\rightarrow(3,3)}(\frac{x-y}{\sqrt x-\sqrt y})=\frac{0}{0}\)

On further solving the equation, we have:

\(\lim_{(x,y)\rightarrow(3,3)}(\frac{x-y}{\sqrt x-\sqrt y})=\lim_{(x,y)\rightarrow(3,3)}(\frac{(\sqrt x)^2-(\sqrt y)^2}{\sqrt x-\sqrt y})\)

Since,

\(a^2-b^2=(a-b)(a+b)\)

Therefore,

\(\lim_{(x,y)\rightarrow(3,3)}(\frac{(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)}{\sqrt x-\sqrt y})\)

\(\lim_{(x,y)\rightarrow(3,3)}(\sqrt x+\sqrt y)\)

Now putting the limits, we get:

\((\sqrt3+\sqrt3)\)

\(=2\sqrt3\)

The required answer.