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Math Help - Proving that The 0 raised to infinity is zero.

  1. #1
    Junior Member EliteAndoy's Avatar
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    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:

    limxaf(x)=0, limxag(x)=∞, and limxaf(x)^g(x)=∞ where f(x) is a positive function. Thanks in advance.



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

    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, -1<a<1, \lim_{n\to\infty} a^n = 0. 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
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    Junior Member EliteAndoy's Avatar
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    Re: Proving that The 0 raised to infinity is zero.

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

    By the definition of \lim_{x \rightarrow a}f(x)=0, there is a \delta_1>0 such that whenever |x-a|<\delta_1, |f(x)-f(0)|= |f(x)| < \frac{1}{2} (just set \epsilon=\frac{1}{2} in the definition).

    Now let \epsilon>0, and let M be such that \frac{1}{2^M}<\epsilon. By the definition of \lim_{x \rightarrow a}g(x)=\infty, there is a \delta_2>0 such that whenever |x-a|<\delta_2, |g(x)|>M. So let \delta=\min(\delta_1,\delta_2). Then whenever |x-a|<\delta:

    |f(x)^{g(x)}|<\left|\left(\frac{1}{2}\right)^{g(x)  }}\right| (since |f(x)|<\frac{1}{2})

    <\left|\left(\frac{1}{2}\right)^M}\right| (since |g(x)|>M)

    <\epsilon. Therefore \lim_{x\to{a}}f(x)^{g(x)}=0.

    - Hollywood
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    Junior Member EliteAndoy's Avatar
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    Re: Proving that The 0 raised to infinity is zero.

    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
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    Super Member ILikeSerena's Avatar
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    Re: Proving that The 0 raised to infinity is zero.

    Hi EliteAndoy!

    How about:
    \lim_{n \to \infty} 0^n = \lim_{n \to \infty} 0 = 0
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    Junior Member EliteAndoy's Avatar
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    Re: Proving that The 0 raised to infinity is zero.

    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
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    Super Member ILikeSerena's Avatar
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    Re: Proving that The 0 raised to infinity is zero.

    Quote Originally Posted by EliteAndoy View Post
    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 \fbox{\sum} 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.
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