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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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  • $\begingroup$ It's really a matter of semantics. In the situations where these limits are used, one can often assume that the function will be defined in some open neighborhood anyway $\endgroup$ – Omnomnomnom Feb 21 '16 at 23:11
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    $\begingroup$ The function is not defined for $x^2=y^2.$ $\endgroup$ – zhw. Feb 22 '16 at 1:13
  • $\begingroup$ The function $F(x,y)$ is simply $x^2+y^2$. No cases needed. $\endgroup$ – robjohn Sep 8 '16 at 10:30
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I think your calculations for the lim are acurate, but the definition you gave in the box is incomplete.

We have $\text{Dom}(f)=\mathbb{R}^2\setminus \{x_1=x_2 \lor x_1=-x_2\}$

For the existence of limit we should have 2 conditions satisfied:

  • $(0,0)$ being an accumulation point (limit point) of the domain, which it is.

  • $\forall_{\epsilon>0} \exists_{\delta>0}\forall_x\left(\ \left[x\in \text{Dom}(f) \land (x_1,x_2) \in B_{\delta}(0,0) \right]\Rightarrow f(x_1,x_2)\in B_{\epsilon}(f(0,0)\ \right)$.

Since, $\{x_1=x_2 \lor x_1=-x_2\}$ does not belong to the domain, the intersection of the domain and the ball gives $B_{\delta}(0,0) \setminus \{x_1=x_2 \lor x_1=-x_2\}$, and thus the limit does exist.

The function $F$ you later define, it's the prolongation/extension by continuity of the function $f$, where not only the limit exists but it also takes its value on that point.

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