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

Find the limit and discuss the continuity of the function. $\underset{\left(x,y\right)\to \left(3,1\right)}{lim}\left({x}^{2}-2y\right)$
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Rewiew the given function
$f\left(x,y\right)=\left({x}^{2}-2y\right)$
Objective is to find the limit when $\left(x,y\right)⇒\left(3,1\right)$ and discuss the continuity of the function.
To find the limit when $\left(x,y\right)⇒\left(3,1\right)$, solve
$=\underset{\left(x,y\right)\to \left(3,1\right)}{lim}\left({x}^{2}-2y\right)$
Since the above limit is not forming any indeterminate form, so evaluate the just by substituting the value of x and y.
$=\left({3}^{2}-2\cdot 1\right)$
$=9-2$
$=7$
To check the continuity of the function
Any function f(x,y) is a continuous point $\left({x}_{0},{y}_{0}\right)$ if $\underset{\left(x,y\right)\to \left({x}_{0},{y}_{0}\right)}{lim}f\left(x,y\right)=f\left({x}_{0},{y}_{0}\right)$
Here
$\underset{\left(x,y\right)\to \left({x}_{0},{y}_{0}\right)}{lim}f\left(x,y\right)=f\left({x}_{0},{y}_{0}\right)=\underset{\left(x,y\right)\to \left(3,1\right)}{lim}{x}^{2}-2y=7$ and $f\left({x}_{0},{y}_{0}\right)=f\left(3,1\right)={3}^{2}-2\cdot 1=7$
Since,
$\underset{\left(x,y\right)\to \left({x}_{0},{y}_{0}\right)}{lim}f\left(x,y\right)=f\left({x}_{0},{y}_{0}\right)=7$
So, the given function is continuous at point (3,1).
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