calculus: 0^∞ = ?

Question

0^∞ = ?

Answer 1

0^∞ = 0

The only indeterminate forms are as follows:

0/0

∞/∞

(∞)0

0^0

∞^0

And:

1^∞

If it's not any of those, than it has a real solution.

Answer 2

Despite what I thought at first, 0^∞ is NOT an indeterminate form.

If you plug in to a limit and get 0^∞, you may ALWAYS conclude that the limit is zero.

If, in a^b, a->0 and b->∞, then eventually a will be very, very close to zero; and the fact that b is huge will make a^b even closer to zero than a is (because multiplying a small number by itself many times results in a REALLY small number). So the limit will always be zero.

Answer 3

As mentioned, 0^∞=0 in any sensible context. More precisely, in the context of limits, [(x, y)→(0, ∞)]lim x^y = 0, since for every ε>0, there is an open neighborhood D of (0, ∞) (for instance, the neighborhood given by |x|<ε, y>1) such that if (x, y)∈D, |x^y|<ε. This makes it a determinate form.

The indeterminate forms of exponentiation are ∞^0, 0^0, and 1^∞. 0^∞ is, as mentioned, a determinate form.

Answer 4

0^∞ =0.
This is not one of the indeterminate cases which can be resolved by L'Hospital/s Rule or any other method.The indeterminate cases are 0/0, ∞/∞ , 0*∞ , ∞-∞ , 0^0, ∞^0, and 1^∞ .

lim 0^n = 0
n --> ∞

Answer 5

If you have f=>0 and g=> infinity f^g this is NOT an indeterminated form as if you write it as
e^(g*ln f= the exponent => infinity.
As ln f =>-infinity al depends on g.So if g=>
+infinity the exponent => -infinity and the limit is 0
If g=> to - infinity the exponemnt tends to +infinity and the limit is + infinity.

Answer 6

0 ^inf ===> 0 always

The only indeterminate form is 1^inf

:)

Answer 7

That is an indeterminate form, so you would use methods to transform the limit into something we can use L'Hospital's rule on.

Answer 8

indeterminate

eg (1+x/n) ^n --> exp(n) as n->oo
0^n --> 0 as n -->oo
2^n -->oo as n-->oo