3

Why did $∞^∞$ is not an indeterminate form ? We have seven indeterminate form $$0/0 $$ $$∞/∞$$ $$0\cdot∞$$ $$∞-∞$$ $$0^0$$ $$1^{\infty}$$ $$∞^0$$ but it does not have $$∞^∞ $$why

Adi Dani
  • 16,949
Pritul Dave
  • 218
  • A large number to the power of a large number is always a large number. This is unlike how a small number divided by a small number acts which could either be small, large, or somewhere inbetween. – JMoravitz Oct 11 '17 at 16:19

3 Answers3

3

Because you can definitely determine what it does: it goes to $\infty$.

Randall
  • 18,542
  • 2
  • 24
  • 47
1

So long as $a>1$ we know that $\lim_\limits{x\to\infty}x^a \to \infty$ and that $\lim_\limits{x\to\infty}a^x \to\infty$.

Thus we can easily predict that $\lim_\limits{x\to\infty}x^x \to \infty$ as well. As such, $\infty^\infty$ is very much determinate.

Alec
  • 4,094
  • 2
    $a>1$ for the second one. – Randall Oct 11 '17 at 16:26
  • 1
    I don't know what you mean exactly with what you've written, but $\infty^\infty=\infty$ is meant to be a sorthand for the statement "for any $f,g$, if $\lim_{x\to\infty} f(x)=\infty$ and $\lim_{x\to\infty} g(x)=\infty$, then $\lim_{x\to\infty} (f(x))^{g(x)}=\infty$" –  Oct 11 '17 at 16:26
  • @Randall - Ah, thanks. You're absolutely right. – Alec Oct 11 '17 at 16:27
1

if you admit that $+\infty \times +\infty=+\infty$

$x^x=e^{lnx^x}=e^{x\ln x}$ as $x\ln x\underset{x\to +\infty}{\longrightarrow}+\infty\qquad (+\infty \times +\infty)$

Thus $x^x=e^{x\ln x}\underset{x\to +\infty}{\longrightarrow}+\infty$

Stu
  • 1,690