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Well this format of a limit $0^0$ is an indeterminate form.

I claim that whatever this limit is (which depends on the exact question) should always be in between $[0,1]$.

Is my claim correct?

I have no mathematical proof for it but just a basic idea, that any number base $0$ should try to pull towards $0$, whereas any number power $0$ should pull towards $1$.

Deltaspace
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excuse_dont_exist
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    Have you read the wikipedia article on it? The expression is either left undefined/indeterminate or 1 depending on context. I don't think it is taken to be 0 in any context. – Prasun Biswas Dec 06 '20 at 05:17
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    @PrasunBiswas What about (from that article) $$\lim_{x \to 0^+} \left(e^{-1/x^2}\right)^x?$$ – Théophile Dec 06 '20 at 05:22
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    @Théophile: Ah, my bad. I'm guilty of having just skimmed through it and not seeing those examples. Right, it is best to say "indeterminate" because you can make it to be any non-negative real number or make it to diverge. – Prasun Biswas Dec 06 '20 at 05:27
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    @PrasunBiswas Right, without context, "$0^0$" can be said to be indeterminate, whereas a given instance of a limit approaching $0^0$ could be just about anything. – Théophile Dec 06 '20 at 05:29
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    The flaw in your reasoning is that while base zero will pull the result towards zero when the exponent is positive, it will pull the result towards infinity when the exponent is negative. This is why any result between zero and infinity is possible. – Toby Bartels Dec 06 '20 at 05:37
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    I understand where your claim is coming from, because a lot of staple textbook problems do have those answers between 0 and 1, but of course in general that does not have to be true. Apart from the answers below, in some cases the limit does not even exist! – imranfat Dec 06 '20 at 06:18

2 Answers2

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Your claim is wrong. You can choose any nonnegative number as the limit: $$\lim_{x\to0^+}\left(e^{-1/x}\right)^{ax} = e^{-a}.$$

This example is from the Wikipedia article on $0^0$, as @PrasunBiswas suggested. There are other examples there showing a limit of infinity, etc.

Théophile
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  • You can also get $0$ with $(e^{-1/x^4})^{ax^2}$, or $\infty$ with $(e^{-1/x^4})^{-ax^2}$, with $a>0$. (These exponents achieve $2$-sided limits, but can be reduced if we only need $x\to0^+$.) – J.G. Dec 06 '20 at 07:36
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That's not true in general. Take $$y(t)=e^{-1/t}$$ and $$x(t)=-5t$$ say. Then $$\lim_{t \rightarrow 0^+} y(t) = \lim_{t \rightarrow 0^+} x(t)=0,$$ but $$\lim_{t \rightarrow 0^+} (e^{-1/t})^{-5t}= e^5 >1.$$

For an even more dramatic example take $y(t)=e^{-1/t^2}$ and $x(t)=-t$.

HOWEVER, if $y(t)$ and $x(t)$ as above are both positive for then infact the limit will indeed fall between 0 and 1, as any positive number less than 1 raised to a positive power is in [0,1].

Mike
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