Thread: Proving that The 0 raised to infinity is zero.

Hi, we are asked to prove that 0 raised to infinity is zero by using limit definitions such as:

limx→af(x)=0, limx→ag(x)=∞, and limx→af(x)^g(x)=∞ where f(x) is a positive function. Thanks in advance.

Since infinity is a symbol and not a number, there needs to be a definition for what you mean by taking 0 (or any other number) to the infinity power.

I think what you mean to say is that you want to prove that for any real number a, , . This should be pretty easy if you've done proofs about limits before. If not, you can just think about what happens if you keep multiplying over and over by a.

- Hollywood

Unfortunately though, we are only asked to prove that 0 raised to infinity is equal to zero and is not of indeterminate form. But we are given a hint to use these limit definitions:
\lim_{x \rightarrow a}f(x)=0, \lim_{x \rightarrow a}g(x)=\infty, and then evaluating \lim_{x \rightarrow a}f(x)^{g(x)}. I really wish it was more specific and yeah, it would be pretty easy if it was the same case as -1<a<1, \lim_{n \rightarrow \infty}a^n=0.
-Andoy

By the definition of , there is a such that whenever , (just set in the definition).

Now let , and let M be such that . By the definition of , there is a such that whenever , . So let . Then whenever :

(since )

(since )

. Therefore .

- Hollywood

Now there's where I think I screwed up before I posted this thread. I tried to prove it using the same method but without setting a constant for my Epsilon. I'm not really sure if I'm allowed to set my Epsilon to a constant but if I was, then you sure did help me lots. Thanks.
-Andoy

Hi EliteAndoy!

How about:

Hi Serena! That was a good idea, and by logic, we can say that 0 raised to infinity is technically 0 raised to a very large finite number since infinity is not a number. But unfortunately, the limit you provided just now do not prove anything that in a function k^n where k is approaching zero and n approaches infinity that the whole function itself would go to zero. Nice try though. What I did is I used the same proof of hollywood but I defined abs(f(x)-L)<Epsilon, as abs(f(x))<epsilon since L is defined zero in the limit as x approaches a. After that I used the squeeze theorem to basically say that f(x) approximates to (-E,E). From there on, I took the limit to infinity of approx f(x) as E because f(x) is assumed to be positive. From there I proved that Limit of f(x)^g(x) as x approaches infinity is zero. Thanks for reply guys, and feel free to say if I got anything wrong from my proof.

P.S. how do you guys write those neat equations right there? I tried latex but It doesn't seem to work. XD

Originally Posted by EliteAndoy

Hi Serena! That was a good idea, and by logic, we can say that 0 raised to infinity is technically 0 raised to a very large finite number since infinity is not a number. But unfortunately, the limit you provided just now do not prove anything that in a function k^n where k is approaching zero and n approaches infinity that the whole function itself would go to zero.

Oh? I didn't see in your problem statement that you had to use a function k^n.
Seems like overcomplicating the problem.

P.S. how do you guys write those neat equations right there? I tried latex but It doesn't seem to work. XD

If you click Go Advanced, type a latex formula, select it, and click the button, it'll become Latex.
Alternatively, just type [ TEX ] <your formula> [ /TEX ]. Without the spaces of course.
Or if you click Reply With Quote, you can see how someone else did it.