Is a consistent raised to the strength of infinity indeterminate? ns am simply curious. Say, because that instance, is $0^infty$ indeterminate? Or is it just 1 elevated to the infinity the is?



$egingroup$ If we form those expressions right into altoalsimce.orgematica, however, it tells united state that 0^infinity is 0 and 1^infinity is indeterminate. $endgroup$
No, the is zero.

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Consider the function $f(x,y) = x^y$ and also consider any sequences $(x_0, y_0), (x_1, y_1), ldots$ with $x_i o 0$ and $y_i o infty$. It is basic to check out that $f(x_n,y_n)$ converges come zero: permit $epsilon > 0$. For part $N$, $|x_i| 1$ for all $i geq N$, for this reason $|f(x_i,y_i)| 1$, oscillates without converging for $c leq -1$, and is indeterminate once $c=1$.


$egingroup$ due to the fact that the question is ambiguous: altoalsimce.orgematica, together it implies, interprets it as a limit of the complex exponential. This is really different 보다 user7530's interpretation together an procedure on the prolonged real numbers, where the limit supplied to compute it is limited to optimistic bases and real exponents. $endgroup$
Since the question pertained to the front web page again, let"s execute a complicated example, to display that (at least this time) altoalsimce.orgematica is no crazy...

For $t>0$, permit $f(t) = t, g(t) = i/t$. Then$$lim_t o 0+ f(t) = 0,qquad lim_t o 0+g(t) = infty\f(t)^g(t) = expfraci;log tt$$and $altoalsimce.orgrmRe Big(f(t)^g(t)Big)$ (pictured) does not converge together $t o 0^+$.



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just how to prove the a kind like $0^infty$ is not an indeterminate form? What makes a type indeterminate?
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