# exponential function or any number raised to the power infinity [closed]

e raised to the power 1 (e^1)

e raised to the power infinity (e^infinity)

e raised to the power of minus infinity (e^-infinity)

0 raised to the power infinity (0^infinity)

0 raised to the power -infinity (0^-infinity)

any number like 2 or 3 raised to the power of plus infinity or minus infinity

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## closed as off-topic by 900 sit-ups a day, RecklessReckoner, Tomás, Potato, HakimJul 19 at 23:23

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Point no. 3 in the answer has one small mistake - its x to the power -n. –  user66721 Mar 14 '13 at 11:49

Remark 1: You should care more about the quality of your question.

Remark 2: There are more rigorous argument for some of these, but I stress on the intuition since you seem clueless. For more information you should read in the Wikipedia page on exponentiation and the math.SE post titled any number raised to the power of infinity and related post titled Negative 1 to the power of Infinity.

1. $e^1 = e$. For all $x$ we have $x^1 = x.$

2. $e^{\infty} = \infty$. For all $x > 1$ we have $\lim\limits_{n \to \infty} x^n = \infty$. Think of it in terms of a simple example $2^1,$ $2^2,$ $2^3,$ and so on. It gets bigger and bigger.

3. $e^{-\infty} = \dfrac{1}{e^\infty} = 0$. For all $0 < x < 1$ we have $\lim\limits_{n \to \infty} x^{n} = 0$. Think of it in terms of a simple example $\frac{1}{2^1}=\frac{1}{2},$ $\frac{1}{2^2}=\frac{1}{4},$ $\frac{1}{2^3}=\frac{1}{8},$ and so on. It gets smaller and smaller.

4. $2^{\pm \infty}$ and $3^{\pm \infty}$ are just similar to $e^{\pm \infty}$.

5. $0^\infty = 0$. No matter how many times you multiply $0$ together, you'll get $0$.

6. $0^{-\infty}$ is undefined. For all $x$ we have $x^{-n} = \dfrac{1}{x^n}$. For all $n > 0$ we have $0^n = 0$. And finally $\dfrac{1}{0}$ is undefined. So putting it together: $0^{-\infty} = \dfrac{1}{0^\infty} = \dfrac{1}{0} = \text{undefined}$.

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## protected by Zev ChonolesJun 22 '13 at 16:14

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