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[lemon10]

It would be 0.

[Pnx]

Thought so, thanks.

[Another]

There are different styles of limits. If your definition says that the point itself is included in the set of points that shrinks, then the limit would be undefined. Most common definition excludes the point itself, but check precise formulation in your textbook.

[MagmaMcFry]

To be fair, limits wouldn't make any sense at all if the value of the limit input itself were included.

[Another]

That choice simplifies definition of a continuous function.

[MagmaMcFry]

Well, the definitions of continuosity would be "The limit exists everywhere" versus "The limit value is equal to the function value everywhere". Not that big of a deal, really.

[ZetaX]

I think the actual reason for normally not including the point is didactical: It is way easier to tell people something like "as it gets closer to a, the values get closer to f(a)" rather than "if the limit exists, some two sequences have to converge to the same, therefore ...".
Apart from that there should not be much need for it to exclude the point as long as the function is already defined there (unlike, e.g., the differential quotient). All cases I can think of right now and where you want to exclude the point on purpose do already have a special name (e.g. right sided limits).

[Virex]

If you include the value in the limit, then the limit of, for example


y = 7 if x = 1
y = 3x if x != 1




would be 7 instead of the value of 3 that one would expect.

[MagmaMcFry]

No, the limit would simply be undefined.

[Tarqiup Inua]

No, the limit would simply be undefined.

Actually, the limit would be simply 3, but I suppose that was just an oversight on your part :-)

The function wouldn't be continuous because at the point x=3 exactly because 3 != 7, but the limit would be exactly what would you expect.

(Eh, sorry, I understand what you guys might be onto but you have to understand - there is no "if you include the value of function at the point into the limit", that's simply nonsense... it's like saying "if you add guitar into gravitational constant" - the value itself has nothing to do with it, that's the point of limit! It would be entirely useless if it said the same thing as value of function)

[MagmaMcFry]

No, the limit would simply be undefined.

Actually, the limit would be simply 3, but I suppose that was just an oversight on your part :-)

The function wouldn't be continuous because at the point x=3 exactly because 3 != 7, but the limit would be exactly what would you expect.

Actually, this discussion is about a non-standard definition of limits where f(3) is not ignored when approaching x=3. Don't you ever accuse me of overlooking anything. 

[ZetaX]

The definition of a limit of a sequence is well defined and does not suffer this kind of dilemma. The definition of a limit of a function (when approaching[and maybe equaling] a point) is less unique as it talks about "all" sequences converging to that point, where "all" depends on the case:
a) any
b) those that omit the point (the usual one)
c) those that are greater (right sided limit)
d) ... smaller (left ...)
e) only linear sequences
etc.

[Pnx]

Here's one for you:

Find the limit as x approaches infinity of (x+1)^x / (x^x)

This apparently equals e, which I know is equal to (1+(1/x))^x however I'm having trouble figuring this out algebraically.

[Jim Groovester]

(1 + 1/x)^x = (x/x + 1/x)^x = ((x+1)/x)^x = (x+1)^x/x^x

[Mego]

Latest problem: How do you find a 3-variable parametric function (x(t), y(t), z(t)) for the rotation of a 2-variable rectangular function about an axis (y=x^2 around the x-axis, for example)?