Why does any number to the power of negative infinity equal zero?

Answer 1

The question is misleading. Infinity and negative infinity are not numbers, so you cannot raise them to powers, nor can you raise numbers to them as powers.

On the other hand, you can approach infinity: since ax = ex ln(a), that means limx->-∞ ax = limx-> -∞ ex ln(a). And as x -> -∞, cx -> -∞ for any constant c. And as the exponent of ex goes to minus infinity, the function itself goes to zero. So it all hangs on that last fact, that limx-> -∞ ex = 0, a fact that is not proven in elementary calculus, but in a more advanced course, and only accepted until then.

To prove it, remember the definition of limit. The statement limx-> -∞ ex = 0 means that if we want to get ex within 1 unit of 0, or within 0.1 of 0, or 0.01, or 0.001, or even 0.000 001 of 0, we just need to pick the right x - anything smaller than a certain number will do. Thus, to get ex within 1 unit of 0, you just need to get ex < 1. Taking the log of both sides, x < ln 1 = 0, so anything negative for x will produce the desired results. For other values, you do the same thing: thus, to get ex within 0.01 of 0, we need ex < 0.01, so x < ln 0.01, or x < -2 ln 10, so anything less than -2 ln 10 will do. To get ex within 0.000 001 of 0, we need ex < 0.000 001, or x < ln 0.000 001 = -6 ln 10, and so on forever.

I would like to put it like this: 1/ x^infinity = 0 (x>1) if true then the denominator is tending to infinity and then 1 = 0 * infinity, which is untrue