Consider the limit $$\lim_{(x,y)\to (0,0)} \frac{x^4-y^4}{x^2- y^2}$$
Now, many would argue that:
$$\lim_{(x,y)\to (0,0)} \frac{x^4-y^4}{x^2- y^2} = \lim_{(x,y)\to (0,0)} \frac{( x^2-y^2)( x^2+y^2)}{ x^2-y^2}$$
$$\lim_{(x,y)\to (0,0)} \frac{x^4-y^4}{x^2- y^2} = \lim_{(x,y)\to (0,0)} (x^2+y^2) = 0$$
Yet, when I read the (or some, in some book) definition it says
if $f(x,y)$ is a real function defined at every point in an open disk
containing $(a,b)$ excluding the point $(a,b)$ etc $\epsilon$... etc $\delta$..
If we accept the above definition, or something similar, then the above limit does not exist. However if we define a new function curing the issue, then we have a limit.. such as $$F(x,y)=
\begin{cases}
\frac{x^4-y^4}{x^2- y^2} &\text{if } x^2\ne y^2 ,\\
x^2+ y^2 &\text{if } x^2=y^2
\end{cases}$$
Then the limit for this function is ok. I just want to make sure I am not missing something.
I ask because calculus professors often teach students that the limits must exists and be equal along "every" path to hope for a limit. This breaks down along the path $y=x$.
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