# Indeterminate forms

1. Apr 16, 2005

### 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

$$a^\infty$$

$$1^\infty$$

2. Apr 16, 2005

### 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.

3. Apr 16, 2005

### RadiationX

yes i don't know why these are true

4. Apr 16, 2005

### dextercioby

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

Daniel.

5. Apr 16, 2005

### 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?

6. Apr 16, 2005

### dextercioby

Where's the 0...?

Daniel.

7. Apr 16, 2005

### 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.

8. Apr 16, 2005

### RadiationX

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

9. Apr 16, 2005

### dextercioby

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

Daniel.

10. Apr 16, 2005

### RadiationX

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

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