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 [tex]e^{x} [/itex].U'll see where the first 2 come from.As for the 3-rd,i think it's an "okay" operation in $$\bar{\mathbb{R}}$$.<br /> <br /> Daniel.[/tex]

 

[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.