Question

find the limit.
\\( \lim _ { ( x, y ) \
ightarrow ( 9,9 ) } \left( \frac { x ^ { 2 } y ^ { 3 } - x ^ { 3 } y ^ { 2 } } { x ^ { 2 } - y ^ { 2 } } \
ight) \\)

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(x^{2}y^{3}-x^{3}y^{2}\) and the denominator \(x^{2}-y^{2}\).
The numerator can be factored as \(x^{2}y^{2}(y - x)\) (since \(x^{2}y^{3}-x^{3}y^{2}=x^{2}y^{2}(y - x)\)) and the denominator is a difference of squares, so \(x^{2}-y^{2}=(x - y)(x + y)=-(y - x)(x + y)\).
So the expression becomes:

$$ \frac{x^{2}y^{2}(y - x)}{-(y - x)(x + y)} $$

Step2: Cancel common factors

We can cancel out the common factor \((y - x)\) (assuming \(y
eq x\), and since we are taking the limit as \((x,y)\to(9,9)\), we can consider the neighborhood around \((9,9)\) where \(y
eq x\) except at the point itself, and the limit is about the behavior near the point). After canceling, we get:

$$ \frac{-x^{2}y^{2}}{x + y} $$

Step3: Substitute the limit values

Now, substitute \(x = 9\) and \(y = 9\) into the simplified expression \(\frac{-x^{2}y^{2}}{x + y}\).
First, calculate \(x^{2}=9^{2} = 81\), \(y^{2}=9^{2}=81\), and \(x + y=9 + 9 = 18\).
Then the numerator is \(-81\times81=-6561\) and the denominator is \(18\).
So \(\frac{-6561}{18}=-\frac{729}{2}\) (dividing numerator and denominator by 9: \(\frac{-6561\div9}{18\div9}=\frac{-729}{2}\)).

Answer:

\(-\frac{729}{2}\)