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Why is it that e raised to the power of negative infinity would equal 0 instead of negative infinity? I am working on problems with regards to limits of integration, specifically improper integrals and a little confused as to what things approach infinity our negative infinity versus approaching zero

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    $\begingroup$ "Negative infinity" is not a number, so "e raised to the power of negative infinity" is nothing. What you want to know is «why is $\lim\limits_{x\to-\infty}e^x=0$?». $\endgroup$ – Mariano Suárez-Álvarez Mar 16 '15 at 3:05
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    $\begingroup$ The principle root of a positive number raised to any real power (positive or negative) is positive. Consider the phrase more as $\lim\limits_{n\to -\infty} e^{n} = \lim\limits_{n\to\infty} e^{-n} = \lim\limits_{n\to\infty}\frac{1}{e^n}$. You should know that $e$ raised to a "large" number is "large", and 1 divided by a "large" number is very near zero. $\endgroup$ – JMoravitz Mar 16 '15 at 3:07
  • $\begingroup$ Ok, because I was thinking of it like a zero, e raised to zero =1, so e raised to negative infinity = nothing $\endgroup$ – apple.pi Mar 16 '15 at 3:08
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    $\begingroup$ @MarianoSuárez-Alvarez has said exactly what you need. Another way to see this is to look at the graph of $y=e^{x}$. Isn't there a horizontal asymptote towards the left (i.e., as $x\to-\infty$, doesn't $y$ settle?)? $\endgroup$ – MPW Mar 16 '15 at 3:09
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    $\begingroup$ $e^{-\infty}=\dfrac1{e^\infty}=\dfrac1{\infty}=0$. $\endgroup$ – Lucian Mar 16 '15 at 4:46
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Note that: $e^{k} = \dfrac{1}{e^{-k}}$. Thus if $k \to -\infty \Rightarrow -k \to +\infty$, and $e^{-k} \to +\infty$ because $e^{-k} > -k$, and the as the latter approaches $+\infty$, so does the former, hence $e^k=\dfrac{1}{e^{-k}} \to 0$.

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First of all: $-\infty$ is not a number and you should always be cautious when you use it as if it were one. This means you should not read $e^{-\infty}$ as an exponentiation of two real numbers, but as the question

If I compute $e^{-N}$ for really really big values of $N$, what does my result look like (approach to)?

Mathematicians write this as the question for the value $\lim_{x\to\infty}e^{-x}$. You probably heard that

$$e^{-N}=\frac1{e^N}$$

and that $e^N$ is enormously bigger than $N$ for large values of $N$. This means that $1/e^N$ is very very small, i.e. close to zero. And it comes closer to zero if we make $N$ larger. At no point it crosses the zero. This brings us to the point that $e^{-\infty}$ is considered zero because

  1. no bigger number makes sense. E.g. assume $e^{-\infty}=0.0001$. But already $e^{-10}<0.0001$. It makes no sense to make $e^{-\infty}$ bigger than that because bigger values of $N$ should make it smaller. And you will always find a way to make $e^{-N}$ smaller if you do not choose $e^{-\infty}=0$.
  2. no negative number makes sense because this would be some kind of a jump in the function. We are always above zero and suddenly we should be below? Sounds strange for a nice (say continuous) function like $e^x$.

Also if you look at the graph of $e^{-x}$ you see that it approaches $0$ as $x\to\infty$.

Additionaly $e^x$ is known for its property of being positive everywhere (on the real line). This does not mean that $e^{-\infty}=0$, but it makes the choice $e^{-\infty}=-\infty$ very unnatural.

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    $\begingroup$ Good explanation of limit as n->infinite to a beginner! $\endgroup$ – richard1941 May 2 '18 at 0:58
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$$e^{-k} = \dfrac{1}{e^{k}}$$ Also $e^{\infty} = \infty$ and $e^{-k}= \dfrac{1}{e^k} \Rightarrow e^{-\infty} = \dfrac{1}{\infty}= 0$

(anything divided by infinity is zero)

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    $\begingroup$ It is not rigorous. The fact that the question is sketchy does not justify giving a sketchy answer. $\endgroup$ – wroobell Sep 13 '16 at 20:10

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