Find the limit and discuss the continuity of the function. lim_{(x,y)rightarrow(3,1)}(x^2-2y)

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
$=\underset{(x,y)\to (3,1)}{lim}(x^2-2y)$
Since the above limit is not forming any indeterminate form, so evaluate the just by substituting the value of x and y.
$=(3^2-2\cdot 1)$
$=9-2$
$=7$
To check the continuity of the function
Any function f(x,y) is a continuous point $(x_0,y_0)$ if $\underset{(x,y)\to (x_0,y_0)}{lim}f(x,y)=f(x_0,y_0)$
Here
$\underset{(x,y)\to (x_0,y_0)}{lim}f(x,y)=f(x_0,y_0)=\underset{(x,y)\to (3,1)}{lim}{x}^2-2y=7$ and $f(x_0,y_0)=f(3,1)=3^2-2\cdot 1=7$
Since,
$\underset{(x,y)\to (x_0,y_0)}{lim}f(x,y)=f(x_0,y_0)=7$
So, the given function is continuous at point (3,1).