- #1
Schniz2
- 19
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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?
Schniz2 said: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?
Schniz2 said: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]
CompuChip said: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).
The limit of infinity over its square root is equal to infinity. This means that as the number being divided by the square root gets larger, the result will also get larger and approach infinity.
To solve for the limit of infinity over its square root, you can use the L'Hopital's rule or the limit definition of a derivative. This involves taking the derivative of both the numerator and denominator and then evaluating the limit as the value approaches infinity.
The limit of infinity over its square root represents the rate at which a number grows as it approaches infinity. It can also be interpreted as the maximum possible value that the expression can reach.
No, the limit of infinity over its square root cannot be negative. Since infinity is always a positive value, dividing by its square root will also result in a positive value. The limit can approach negative infinity, but it cannot be negative itself.
The limit of infinity over its square root is used in various fields of science and mathematics, such as in calculus, physics, and engineering. It can help in understanding the growth rate of certain phenomena and in solving complex equations and problems.