I was thinking of a strange case in which we can say the limit is undefined. ( I mean different than the limit does not exist.)
So I generated the graph of a piecewise function $f(x)$. There is a discontinuity but the graph is is of the same function just in case.
$$f(x)=\begin{cases} 1 & -5<x<-4 \\x^2 & x>0 \end{cases}$$
Off course the limit does not looks to exist( or perhaps it does).
Why it is so?
See below!
$$\lim_{x \to \ -2^-} f(x)= \text{the limit does not exist}$$
Why did I say that?
Because I think that as we are moving closer to -2 from left , the value of $f(x)$ seems to approach undefined.
And if I were to create a table also. I reach this conclusion but it doesn't makes sense because I cannot say the if there is a x in the range of $\epsilon$ of undefined we can find a $\delta$ and the formal limit definition. So according this it brings me to the conclusion the limit does not exist.
But the problem with my this conclusion is that, I never imagined that an one sided limit cannot exist.
So I doubt if I am correct. If not why?
Now,
$$\lim_{x \to \ -2^+} f(x)= \text{the limit does not exist}$$
And the reason is same again.
Now I ask myself:- $$\lim_{x \to \ -2} f(x)= ?$$
Off course the limit does not exist but we see the left and right limit , both were equal right so shouldn't exist?
Or
We cannot say they were equal because the answers were not numbers so we cannot compare and say they are equal and and hence the limit does not exist.
Which one is correct?
The prerequisite for a limit to exit is that the function should be well defined in the neighbourhood of where we will be finding the limit.In this case the function is not defined at or near x=2 so limit does not exist.It is not that the value of () seems approaches undefined, instead the limit does not exist as the function is well defined only in its domain $(-5,-4)\cup(0,\infty)$.