Rewiew the given function

\(f(x,y)=(x^2-2y)\)

Objective is to find the limit when \((x,y)\Rightarrow(3,1)\) and discuss the continuity of the function.

To find the limit when \((x,y)\Rightarrow(3,1)\), solve

\(=\lim_{(x,y)\rightarrow(3,1)}(x^2-2y)\)

Since the above limit is not forming any indeterminate form, so evaluate the just by substituting the value of x and y.

\(=(3^2-2\cdot1)\)

\(=9-2\)

\(=7\)

To check the continuity of the function

Any function f(x,y) is a continuous point \((x_0,y_0)\) if \(\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)=f(x_0,y_0)\)

Here

\(\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)=f(x_0,y_0)=\lim_{(x,y)\rightarrow(3,1)}x^2-2y=7\) and \(f(x_0,y_0)=f(3,1)=3^2-2\cdot1=7\)

Since,

\(\lim_{(x,y)\rightarrow(x_0,y_0)} f(x,y)=f(x_0,y_0)=7\)

So, the given function is continuous at point (3,1).

\(f(x,y)=(x^2-2y)\)

Objective is to find the limit when \((x,y)\Rightarrow(3,1)\) and discuss the continuity of the function.

To find the limit when \((x,y)\Rightarrow(3,1)\), solve

\(=\lim_{(x,y)\rightarrow(3,1)}(x^2-2y)\)

Since the above limit is not forming any indeterminate form, so evaluate the just by substituting the value of x and y.

\(=(3^2-2\cdot1)\)

\(=9-2\)

\(=7\)

To check the continuity of the function

Any function f(x,y) is a continuous point \((x_0,y_0)\) if \(\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)=f(x_0,y_0)\)

Here

\(\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)=f(x_0,y_0)=\lim_{(x,y)\rightarrow(3,1)}x^2-2y=7\) and \(f(x_0,y_0)=f(3,1)=3^2-2\cdot1=7\)

Since,

\(\lim_{(x,y)\rightarrow(x_0,y_0)} f(x,y)=f(x_0,y_0)=7\)

So, the given function is continuous at point (3,1).