Question

help please, I don't fully understand the concept of infinity:

Answer 1

given: \[w=\infty\]
what is:\[Z= \sqrt{R ^{2}+(wL-\frac{ 1 }{ wC})^{2}}\]

Answer 2

First, the short answer is that Z is infinite as well.
Where did you come across this question?

Answer 3

Because in mathematics, treating "infinity" like it is some number is actually considered really sloppy. In actuality, INFINITY is not a number--it's a concept.
Basically, the best way to restate your question is to say that the limit of Z as w goes to infinity is equal to positive infinity.... Is this helpful??

Answer 4

@Splash_Dance sorry for the late reply,
I got this is a physics formula for total impedance Z, we had to evaluate it for w= 0 and w= infinity.

Don't really understand how Z will be equal to infinity..

Answer 5

First, I should clarify something: I absolutely HATE physics.

Second, Physicists are weird: They do things that make Mathematicians cringe (like plugging in "infinity" into a formula).

Nevertheless, if what you're doing is finding the limit of "Z" as "w" goes to infinity, then "Z" definitely goes to infinity as well.

But, to be frank, I can't help you with the physical intuition because--as I indicated earlier--I REALLY hate physics.

What part of the problem are you confused about??

Answer 6

we have the these two terms wL and 1/wC: what happens when we multiplying them with infinity gives infinity?

Answer 7

Don't worry about all that--here, give me a second and I'll rewrite the expression...

Answer 8

ok thank you very much! :)

Answer 9

Ok, so what you're asking is:

\[\lim_{w \rightarrow \infty} Z = \lim_{w \rightarrow \infty} \sqrt{R ^{2} + (wL - \frac{ 1 }{ wC })^{2}}\]
\[= \sqrt{\lim_{w \rightarrow \infty}( R ^{2} + (wL - \frac{ 1 }{ wC })^{2}})\]
\[=\sqrt{R ^{2} + \lim_{w \rightarrow \infty} (wL - \frac{ 1 }{ wC })^{2}}\]

Now, when w goes to infinity, the term "1/(wC)" goes to zero--i.e. you can ignore it. And all that other stuff--the square root, R-squared, the square-ing, etc.--none of that matters. [Well, actually, the sign of "L" would've mattered but the squaring ensures that it's positive.]

Quick question: Is this helping?

Answer 10

ok I understand the derivation/simplifications and so as w approaches infinity, (wL)^2 is equal to infinity?
and it is helping a lot

Answer 11

Yeah, pretty much. Except you'd probably want to say it more like, "As w goes to infinity, Z increases without bound". Any other questions?

Answer 12

nope, thank you very much! ^^