Solving Indeterminate Forms: e^∞, √∞, a^∞, 1^∞

[RadiationX]

I'm having trouble recognizing when an expression produces an indeterminate form. for exampe what are the following:

e^\infty

\sqrt{\infty}

more generally what is

a^\infty

1^\infty

 

[dextercioby]

e^{\infty} is something unclear...

e^{-\infty}=0

e^{+\infty}=+\infty

\sqrt{+\infty}=+\infty

As for 1^{\infty} and the asymptotic limit of the general exponential,they require a special analysis...

Daniel.

 

[RadiationX]

e^{\infty} is something unclear...

e^{-\infty}=0

e^{+\infty}=+\infty

\sqrt{+\infty}=+\infty

As for 1^{\infty} and the asymptotic limit of the general exponential,they require a special analysis...

Daniel.

yes i don't know why these are true

 

[dextercioby]

Which ?The ones i wrote...?Take a look at the definition of the exponential function and expecially at the graph of e^{x} [/itex].U&#039;ll see where the first 2 come from.As for the 3-rd,i think it&#039;s an &quot;okay&quot; operation in \bar{\mathbb{R}}.<br /> <br /> Daniel.

 

[RadiationX]

my misunderstanding stems from this problem:

Evaluate: \lim_{x\rightarrow\infty}\frac{\sqrt{x}}{e^x}

i have to use L' Hopital's rule and the above ruduces to this:

\lim_{x\rightarrow\infty}\frac{1}{2\sqrt{x}e^\infty}=0

now isn't 0\infty indeterminate?

 

[dextercioby]

Where's the 0...?


Daniel.

 

[Theelectricchild]

Could he be thinking to split up the limit:

\lim_{x\rightarrow\infty}[(\frac{1}{2\sqrt{x}})(\frac{1}{e^\infty})]=0

but that would give the determinant form of zero times zero, which is undoubtedly zero--- not zero times infinity.

 

[RadiationX]

i didn't know that e^-\infty was not equatl to e^\infty

 

[dextercioby]

Well,that's because u don't know how the graph of e^{x} looks like...


Daniel.

 

[RadiationX]

you are totally correct. i didn't even think about looking at that graph.