What is $ \lim_{(x,y)\to(2,2)}\frac{x^4-y^4}{x^2 - y^2} $? I have limit:
$$
\lim_{(x,y)\to(2,2)}\frac{x^4-y^4}{x^2 - y^2}
$$
Why is the result $8$ ?
 A: Hint:
$$x^4-y^4 = (x^2+y^2)(x^2-y^2)$$
A: $$\lim_{x,y\to{2,2}}\frac{x^4-y^4}{x^2-y^2}$$ $$= \lim_{x,y\to{2,2}}\frac{(x^2+y^2)(x^2-y^2)}{x^2-y^2}$$
$$\lim_{x,y\to{2,2}} x^2+y^2 =...$$
A: I am really surprised (at other answers which have been upvoted) as to how we can think of the limit when the function $f(x, y) = \dfrac{x^{4} - y^{4}}{x^{2} - y^{2}}$ is not defined in a neighborhood of point $(2, 2)$. Note that the definition of $f(x, y)$ assumes that we must not have $x^{2} = y^{2}$ i.e. $x = \pm y$. So clearly in any neighborhood of point $(2, 2)$ we will will have points $(x, y)$ with $x = y$.
The limit of the function therefore does not exist.
Update: Please also check this question and its answer which deals with a similar scenario in single variable.
A: The limit you are looking up to has a 'hole' in its value as $(x,y)\rightarrow(2,2)$. In order to eradicate that, we have to factor out the hole which is quite easy in this case. It can be done as follows:
$$\lim_{(x,y)\rightarrow(2,2)}\dfrac{x^4-y^4}{x^2-y^2}=\lim_{(x,y)\rightarrow(2,2)}\dfrac{(x^2+y^2)(x^2-y^2)}{x^2-y^2}=\lim_{(x,y)\rightarrow(2,2)}x^2+y^2$$  
Now we have successfully removed the 'hole' in the value by getting rid of the $\dfrac00$ format of the expression.  
Now, substituting the values of $x$ and $y$ in the expression gives us,
$$\lim_{(x,y)\rightarrow(2,2)}x^2+y^2=2^2+2^2=4+4=8$$  
