0 in the power of infinite ?

Question

what's the result of it? my teacher said that it 's 0 and wants us to prove it.
note: zero and infinit are limit ones

Answer 1

So you have lim x^y when x->0 and y->+oo (the statement is not true in general if y->-oo). The fact that y->+oo means that there is such value of parameter when all y > M, where M is arbitrary large. The fact that x->0 means that there is such value of parameter when all x < k, wher k is arbitrary small. For our purpose it is enought to know that y > 1, then x^y < x for x < 1. Then 0 <= lim x^y <= lim x = 0.

You can use delta-epsilon definition too.

Answer 2

0 to any positive integer power is 0.

Imagine making a sequence of numbers like this:

0^1, 0^2, 0^3, 0^4, 0^5....

All the terms will be zero. Therefore the limit of the sequence is zero. Thats what we mean when we say "Zero to the inifinte power is zero".

Answer 3

If by zero you mean truly zero, then the proof by induction is duck soup.

First, you assume that zero times any finite number, including zero, is zero.

Then you establish the base case that 0^1 = 0.

Finally, you show that if 0^n = 0 for any counting number (positive integer) n, then 0^n * 0 = 0^(n+1) = 0.

From these last two sentences, you can show that 0^2 = 0, then 0^3 = 0, and so on for any arbitrarily large value of n.

Answer 4

0 to the power of infinity has to be 0. Why? Because any number smaller than 1 to the power of infinity is necessarily 0. 0 is smaller than 1, so the limit of something converging to 0 to the power of infinity is 0.

By the way... if a series is converging to 0 and you take the first power, it is converging to 0. If you take the second power, it is converging to 0 even faster. The higher the constant power, the faster the convergence. So if you are making the power increase as well, you will converge even faster.

PS: what I have given you are hints, not the proof, OK?

Answer 5

Since you're talking about limits...

take a very small number, say 1 x 10^ -10 (0,000000001)


now consider this:
(1 x 10^ -10)^2 . you'll get 1 x 10^ -20 (or 0,0000000000000000001)

and
(1 x 10^ -10)^3 . you'll get 1 x 10^ -30 (or 0,00000000000000000000000000001)

you can clearly see that it's going towards 0 (fast). The larger the exponent, the closer you are getting to 0.

don't know if your teacher will consider this as "proof", but it's kinda hard to argue against it.

Answer 6

Sylvainco has the right idea. For small x and large y, first show that for any x1, x2, and y, where x2 < x1, x2^y < x1^y. Then show that for any x, y1, and y2, where y2 > y1, x^y2 < x^y1. Since all of this involves only positive reals, the limiting case 0^oo would be 0, because otherwise any other limit postive real value would lead to a contradiction.

Answer 7

I thought it would be an undefined term such as 0/0. I am not sure there is anything to prove.

Answer 8

it is indefinite in calculus