Why Is 1^∞ an Indeterminate Form in Limits?

[kbaumen]

I'm just curious, why, when solving limits, is $1^\infty$ considered an indeterminate form? Isn't 1 raised to any power equal to 1? Why isn't it so simple?

 

[Tac-Tics]

Probably because if you perturb the 1 at all, the result is either zero or infinite.

 

[╔(σ_σ)╝]

Well that is a very good question but the problem is that it depends on how fast your function is going to 1 or infinity.

Your function can be going so slowly to 1, in which case the limit goes to 0.

It`s the same as the undetermined form 0/0 the function on top and bottom could approach zero at the same speed and the limit could go to 1.

 

[kbaumen]

Well that is a very good question but the problem is that it depends on how fast your function is going to 1 or infinity.

Your function can be going so slowly to 1, in which case the limit goes to 0.

It`s the same as the undetermined form 0/0 the function on top and bottom could approach zero at the same speed and the limit could go to 1.

Ah, yes, the function never actually reaches the "value" $1^\infty$, since we never consider the function at exatcly $x=a$, we are just curious about what it does around that point.

Thank you for your answers.

 

[╔(σ_σ)╝]

Ah, yes, the function never actually reaches the value $1^\infty$, since we never consider the function at exatcly $x=a$, we are just curious about what it does around that point.

Thank you for your answers.

Precisely, you are correct. Even in the precise definition of a limit we only look at the deleted neighborhood of x .

You are welcome.:)