what is the answer.... and why?

[MSI]

* infinity=inf.
questions
a) 1^inf.
b) inf.^0
c) inf.*0

what is the answer for these and why?

[dav2008]

You can't really use any operators with infinity in that sense.

If you have a limit in the form of (1\infty), \infty^0) or (\infty* 0) like you have then that's an indeterminate form and you would have to use another method of finding what that limit is for that specific case.

There are some more indeterminate forms including 0/0 and \frac{\infty}{\infty}

[BobG]

Multiplying by infinity may not be good math technically, but these equations are obviously just trying to make a point.

Doesn't matter how many times you multiply 1 or 0 by itself.

1 x 1 x 1 x 1 x 1 x 1 ..... still equals 1

0 x 0 x 0 x 0 x 0 x 0 ..... still equals 0

By definition, any number to the 0th power equals 1.

[Gokul43201]

* infinity=inf.
questions
a) 1^inf.
b) inf.^0
c) inf.*0

what is the answer for these and why?

The question makes no sense...where did you get it ?

[jcsd]

Multiplying by infinity may not be good math technically, but these equations are obviously just trying to make a point.

Doesn't matter how many times you multiply 1 or 0 by itself.

1 x 1 x 1 x 1 x 1 x 1 ..... still equals 1

0 x 0 x 0 x 0 x 0 x 0 ..... still equals 0

By definition, any number to the 0th power equals 1.

sure, it is obvious that the limit of 1^n as n goes to infinity is 1 and the limit of 0^n as n goes to infinity is 0, but in the abscense of rigourous defintion of infinity which would allow us to evalute 1^inf, etc, it'ds meaninglss to tlak about such things (though 1^inf can be and is used to represent the limit of 1 ^x). 0^0 is not always defined.

[Alkatran]

Rule of thumb: Whenever you're working with infinities, use a limit. There's only one answer that way.

For example:
inf/inf could be seen as:
lim[x->inf] (x/inf) = 0
lim[x->inf] (inf/x) = inf
lim[x->inf] (x/x) = 1

and each answer is just as valid. That's why it's inderteminate.

Applying this to your questions:

a)1^inf.
lim[x->inf](1^x) = 1
OR
lim[x->1(-)](x^inf) = 0
lim[x->1(+)]*x^inf) = inf
b)inf^0
lim[x->inf](x^0) = 1
OR
lim[x->0(-)](inf^x) = 0
lim[x->0(+)](inf^x) = inf
c) inf*0
lim[x->inf](x*0) = 0
OR
lim[x->0(-)](inf*x) = -inf
lim[x->0(+)](inf*x) = inf

Notice that the left and right limits are even different in some cases.
Isn't it wonderful? :surprised