Is a constant raised to the power of infinity indeterminate? I am just curious. Say, for instance, is \$0^infty\$ indeterminate? Or is it only 1 raised to the infinity that is?  \$egingroup\$ If we type those expressions into starrkingschool.netematica, however, it tells us that 0^infinity is 0 and 1^infinity is indeterminate. \$endgroup\$
No, it is zero.

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Consider the function \$f(x,y) = x^y\$ and consider any sequences \$(x_0, y_0), (x_1, y_1), ldots\$ with \$x_i o 0\$ and \$y_i o infty\$. It is easy to see that \$f(x_n,y_n)\$ converges to zero: let \$epsilon > 0\$. For some \$N\$, \$|x_i| 1\$ for all \$i geq N\$, so \$|f(x_i,y_i)| 1\$, oscillates without converging for \$c leq -1\$, and is indeterminate when \$c=1\$. \$egingroup\$ Because the question is ambiguous: starrkingschool.netematica, as it implies, interprets it as a limit of the complex exponential. This is very different than user7530's interpretation as an operation on the extended real numbers, where the limit used to compute it is restricted to positive bases and real exponents. \$endgroup\$
Since the question came to the front page again, let"s do a complex example, to show that (at least this time) starrkingschool.netematica is not crazy...

For \$t>0\$, let \$f(t) = t, g(t) = i/t\$. Then\$\$lim_t o 0+ f(t) = 0,qquad lim_t o 0+g(t) = infty\f(t)^g(t) = expfraci;log tt\$\$and \$starrkingschool.netrmRe Big(f(t)^g(t)Big)\$ (pictured) does not converge as \$t o 0^+\$.  Thanks for contributing an answer to starrkingschool.netematics Stack Exchange!

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