# any number raised to the power of infinity

1) I saw in a book that "the limit as $x$ approaches positive infinity of $e^x$ equals $0$" I want to ask about this?

2) if the $a$ is a negative number and we take a limit like "the limit as $x$ approaches positive infinity of $a^x$ equals?" and if $x$ approaches minus infinity then what happens?

Please also tell me what would happen if $a$ is positive number.

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There's no "a" in the quote you say you saw. – Michael Hardy Oct 5 '11 at 12:27
The limit of $e^x$ as $x$ approaches infinity is $\infty$. Did you mean $\lim_{x\to -\infty}e^x=0$ or $\lim_{x\to \infty}e^{-x}=0$? Also, if $a<0$ the limit $\lim_{x\to\infty}a^x$ is not defined since $a^{\frac{b}{2}}$ does not exist for any $b>0$. In other words even roots of $a$ do not exist. – Joe Johnson 126 Oct 5 '11 at 12:27
i mean limit as x--->- infinity of e to the power x – Zia ur Rahman Oct 5 '11 at 12:34
and i also asked about limit as x--> + infinity of e to the power x – Zia ur Rahman Oct 5 '11 at 12:34
then my question was if we take any number in place of e and do the the same then what happens, if the number is positive and what happens and if the number is negative – Zia ur Rahman Oct 5 '11 at 12:36

As long as the base is greater than one, the same thing happens. $$\lim_{x \to \infty}a^x=\infty, \lim_{x \to -\infty}a^x=0$$ for any $a \gt 1$. $$\lim_{x \to +\infty}a^x=0, \lim_{x \to -\infty}a^x=\infty$$ for any $0 \lt a \lt 1$.
For $a \lt 0$, $a^x$ is undefined in the reals for irrational $x$
@Zia: No. If $a>1$ then $\lim_{x\to\infty}a^x=\infty$ and $\lim_{x\to -\infty}a^x=0$. If $0<a<1$ then $\lim_{x\to\infty}a^x=0$ and $\lim_{x\to -\infty}a^x=\infty$, since $a^{-x}=\frac{1}{a^x}$. – Joe Johnson 126 Oct 5 '11 at 13:21
@ZiaurRahman: I added in the $-\infty$ items – Ross Millikan Oct 5 '11 at 13:31