- #1
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Homework Statement
[tex]\frac{\infty}{\sqrt{\infty}}[/tex]
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?
Homework Statement
[tex]\frac{\infty}{\sqrt{\infty}}[/tex]
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?
To be more precise, I am trying to to a Bode diagram for a simple circuit. I need to estimate |H(j[tex]\omega[/tex])| in decibels as [tex]\omega[/tex] -> [tex]\infty[/tex] and as [tex]\omega[/tex] -> 0.
The transfer function is |H(j[tex]\omega[/tex])| = [tex]\frac{\sqrt{\left(RCj[tex]\omega[/tex]\right)^{2}}}{\sqrt{\left(RCj\omega\right)^{2}+1}}[/tex]
Assuming that [itex]j^2 = -1[/itex], you have
[tex]|H(j \omega)| = \frac{ \sqrt{- R^2 C^2 \omega^2} }{ \sqrt{- R^2 C^2 \omega^2 + 1}}[/tex]
I suggest defining [itex]x = - R^2 C^2 \omega^2[/itex], so you get
[tex]\frac{\sqrt{x}}{\sqrt{x + 1}} = \sqrt{\frac{x}{x + 1}}[/tex]
and then the limits [itex]\omega \to 0, \infty[/itex] correspond to [itex]x \to 0, -\infty[/itex] respectively.
Note that even in the "informal" notation of your first post, this gives
[tex]\sqrt{\frac{-\infty}{1 - \infty}}[/tex]
and not
[tex]\frac{\infty}{\sqrt{\infty}}[/tex]
Anyhow, there are nicer tricks to calculate the limit (for example, multiply by (-x)/(-x) inside the square root).