Is the Limit of Infinity Over Its Square Root Equal to 1?

[Schniz2]

Homework Statement

\frac{\infty}{\sqrt{\infty}}


I would have said it would be infinity because infinity would grow a lot faster than its square root wouldn't it?
But my friend swears the limit is equal to 1?

 

[Feldoh]

Was it something like lim x -> infinity of x/x^(1/2)? What you have isn't a limit...

 

[CompuChip]

\frac{\infty}{\sqrt{\infty}}
is a meaningless expression anyhow.

Do you mean you and your friend were debating whether
\lim_{x \to \infty} \frac{x}{\sqrt{x}}
does not exist, or it exists and is equal to 1 ?

 

[sylas]

Homework Statement

\frac{\infty}{\sqrt{\infty}}


I would have said it would be infinity because infinity would grow a lot faster than its square root wouldn't it?
But my friend swears the limit is equal to 1?

That's not quite a limit as expressed just yet. Try:

\lim_{n \rightarrow \infty} ( \frac{n}{\sqrt{n}} )

When it is actually expressed as a limit, then it may be clearer.

Cheers -- sylas

 

[Russell Berty]

Infinity over infinity is not defined as a number. It is considered an indeterminant form.

Here is an informal example: What is Infinity minus infinity? Consider the set of positive integers, {1, 2, 3, 4, 5, ...}
There are infinitely many "objects" in this set. Now, let us take away an infinite number of objects and count what is left.
Remove 1, 2, 3, 4, ... What is left? Nothing. So Infinity minus infinity = 0.
BUT WAIT!
Remove 2, 3, 4, 5, ... What is left? {1}. One thing is still left. So Infinity minus infinity = 1.
BUT WAIT!
Remove all the even numbers, 2, 4, 6, 8, ... What is left? {1, 3, 5, 7, ...} An infinite number, so Infinity minus infinity = infinity!?

Infinity minus infinity is left undefined, it is an indeterminant form, a concept of two competing forces, one increasing without bound and a counter force removing (or decresing the value) without bound. You need to look closer at the 2 forces to see if you can say anything more.

For example as x -> inf, (x - x/3) -> ? It looks like "inf - inf", they are competing forces. x going up, -x/3 trying to bring it back down. We need to look closer. One way is to simplify the expression x - x/3 = 2x/3 now we see 2x/3 keeps increasing.
So, as x -> inf, (x - x/2) -> inf

With division we see the same process. For example, as x -> inf, where does x/x go?
It looks like inf/inf. But, simplifying we get x/x = 1. So as x->inf, x/x -> 1.

On the contrary, as x-> inf, x^2/x -> goes where? inf/inf? But look closer, x^2/x = x, so
as x -> inf, x^2/x -> inf

To answer your question, i would need to see where the inf/squareroot(inf) form is coming from. Try simplifying the expression, that sometimes works as in the above examples.

 

[Schniz2]

To be more precise, I am trying to to a Bode diagram for a simple circuit. I need to estimate |H(j\omega)| in decibels as \omega -> \infty and as \omega -> 0.

The transfer function is |H(j\omega)| = \frac{\sqrt{\left(RCj\omega\right)^{2}}}{\sqrt{\left(RCj\omega\right)^{2}+1}}

 

[Schniz2]

To be more precise, I am trying to to a Bode diagram for a simple circuit. I need to estimate |H(j\omega)| in decibels as \omega -> \infty and as \omega -> 0.

The transfer function is |H(j\omega)| = \frac{\sqrt{\left(RCj\omega\right)^{2}}}{\sqrt{\left(RCj\omega\right)^{2}+1}}

Cool, that worked well. And took forever.

|H(jw)| = sqrt((RCjw)^2) / sqrt((RCjw)^2 +1)

i need to estimate this as w -> infinity...

 

[CompuChip]

Assuming that j^2 = -1, you have
|H(j \omega)| = \frac{ \sqrt{- R^2 C^2 \omega^2} }{ \sqrt{- R^2 C^2 \omega^2 + 1}}

I suggest defining x = - R^2 C^2 \omega^2, so you get
\frac{\sqrt{x}}{\sqrt{x + 1}} = \sqrt{\frac{x}{x + 1}}
and then the limits \omega \to 0, \infty correspond to x \to 0, -\infty respectively.

Note that even in the "informal" notation of your first post, this gives
\sqrt{\frac{-\infty}{1 - \infty}}
and not
\frac{\infty}{\sqrt{\infty}}

Anyhow, there are nicer tricks to calculate the limit (for example, multiply by (-x)/(-x) inside the square root).

 

[Schniz2]

Assuming that j^2 = -1, you have
|H(j \omega)| = \frac{ \sqrt{- R^2 C^2 \omega^2} }{ \sqrt{- R^2 C^2 \omega^2 + 1}}

I suggest defining x = - R^2 C^2 \omega^2, so you get
\frac{\sqrt{x}}{\sqrt{x + 1}} = \sqrt{\frac{x}{x + 1}}
and then the limits \omega \to 0, \infty correspond to x \to 0, -\infty respectively.

Note that even in the "informal" notation of your first post, this gives
\sqrt{\frac{-\infty}{1 - \infty}}
and not
\frac{\infty}{\sqrt{\infty}}

Anyhow, there are nicer tricks to calculate the limit (for example, multiply by (-x)/(-x) inside the square root).

Nice, thanks!