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

Question

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

darkprince14 · Answer

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

darkprince14 · Answer

@radar @RadEn @Preetha

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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