Question

(Limits of a function in three variables)
Prove by definition, that

Answer 1

\[\lim_{(x,y) \rightarrow (2,3)} (3x^{2} + xy - 2y^{2}) = 0\]

Answer 2

@radar @RadEn @Preetha

Answer 3

Let us first prove that
\[
\lim_{(x,y)\to (2,3)} x y =6
\]

Answer 4

\[
| xy - 6|= | xy - 2y + 2y -6|=\\
| y(x-2) + 2(y-3)|\le |y||x-2| + 2|y-3|

\]

Answer 5

Since y is going to be close to 3, we can assume |y|<4,
Let \(\epsilon >0\), take \(\delta =\frac {\epsilon}6\) and suppose
\[
|x-2| <\delta\\
|y-3|< \delta
\]
then\[
| xy - 6|= | xy - 2y + 2y -6|=\\
| y(x-2) + 2(y-3)|\le |y||x-2| + 2|y-3|\le\\
4 \delta + 2 \delta= 6\delta=\epsilon

\]
We are done

Answer 6

The remaining terms are a one variable limit and can be done easily