Limit at point implies defined at point
There is *no limit* to e^(-1/(x^2)) as x approaches 0.
The idiotic mainstream tendency is to transfer the limit to the exponent, that is,
-oo = Lim_{x \to 0} -1/(x^2)
in which case we say there is no limit because -oo is NOT a limit.
Then what do orangutans do? They say:
0=e^(-oo)
treating infinity exactly as if it were a number. I wonder, does the limit operator jump back and forth between the exponent and e as one would in musical chairs? Chuckle.
As I've stated in the past and continue to state, you cannot have a HOLE in a function at a point c in an interval (a,b) if the function is continuous on the interval and has a limit at c. The bogus mainstream calculus NEEDS holes, but even then, it needs a lot more decrees to stay afloat.
Therefore, the function e^(-1/(x^2)) has NO limit at x=0 otherwise it would be continuous at x=0.
And you thought Swiss cheese was holey eh? :-)))
Wolfram computational engine states that the limit is 0. Tsk, tsk.
Holey functions from old:
• Holey functions from old.
Holes here, holes there, holes everywhere:
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