Verified answer

Think of it this way.

0 × ∞ is indeterminate because:

      0 + 0 + 0 + 0 + ....

is really a(n) copies of b(n):

      b(n) + b(n) + ... + b(n)

where a(n)→∞ but b(n)→0.

The reason this happens to be indeterminate is because zero is the ADDITIVE identity. It's impossible to say whether each b(n) will "cancel out to zero" fast enough to reign in the a(n) number of copies of b(n) that keep tacking on.

For 1^∞, you have:

      1 × 1 × 1 × ...

which is really a(n) copies of b(n):

      b(n) × b(n) × b(n) × ... × b(n)

where a(n)→∞ but b(n)→1.

It has to be b(n)→1 because 1 is the MULTIPLICATIVE identity. If b(n)→0 instead, we definitely get zero! You could look at that as:

limit of b(n) × b(n) × b(n) × ... × b(n)

we find b(k)<1 and say that:

b(n)^a(n) < b(k)^a(n)

But if b(k) is less than one, then b(k)^2 is less than b(k). And b(k)^3 is less than b(k)^2.

In fact, b(k) is just a tiny constant, and powering it to the a(n) as a(n) gets to ∞ goes right to zero.

Because b(k)^a(n) goes to zero, so does b(n)^a(n).

Try this with an example function like (1/n)^n as n→∞. You can take k=2 and just look at (1/2)^n instead. Where does that go as n→∞?

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Let's do a few specific examples of 1^∞.

(x+1)^(1/x) as x→0.

That's the definition of the natural base e (approx 2.718...)

On the other hand:

(1-x)^(1/x) as x→0

That's 1/e (approx 0.367...)

But both of these are of the form 1^∞. In fact:

(1+kx)^(1/x)

Is simply e^k. You can get ANY value you want this way. So it surely must be indeterminate.

0^∞ is determinate because if 0 and ∞ are taken to be limits of functions f(x) and g(y), the limit of f(x)^g(y) as x and y approach their respective limiting values will always be the same - zero.

For example, the limit of x^(1/x) as x -> 0 takes the limiting form 0^∞... it is quite obvious that this limit is zero, because once x falls below one, 1/x becomes greater than one, and as a result 0 < x^(1/x) < x; since x -> 0, we must have x^(1/x) -> 0 also.

As another example, consider 0^(1/x) as x -> 0. Here, it is again quite obvious that the limiting form is 0^∞, and that the value of this limit is zero...

The point is, no matter how we set up our limit, we will get zero. Thus 0^∞ is a determinate form, and may be taken as equivalent to zero.

On the other hand, consider the following two limits:

The limit of 1^x as x -> ∞. This is clearly 1.

The limit of (1+1/x)^x as x -> ∞. This gives the definition of the base of the natural logarithm, e, a transcendental number approximately equal to 2.71828183.

Both, however, take the limiting form 1^∞. So it may be said that 1^∞ is indeterminate.

Loosely speaking, it is indeterminate because, like 0/0, there are multiple values that it may take.

Unity multiplied by unity, infinite times is Unity only!!!