Solving Limits: Help Me Find a Limit!

[stefaneli]

Homework Statement

I don't know how to find a limit, and it's bothering me for a few hours now.
Can someone help me?
j - imaginary unit

Homework Equations

\lim_{\rho \to 0}{\frac{\frac{\sqrt{2}}{2}(-1+j)+\rho \exp(j\theta)}{(\frac{\sqrt{2}}{2}(-1+j)+\rho \exp(j\theta))^2+ \sqrt2(\frac{\sqrt{2}}{2}(-1+j)+\rho \exp(j\theta)))+1}}

The Attempt at a Solution

Solution is:
∞ exp( \frac{∏}{4} - \theta)

 

[Dick]

Homework Statement

I don't know how to find a limit, and it's bothering me for a few hours now.
Can someone help me?
j - imaginary unit

Homework Equations

\lim_{\rho \to 0}{\frac{\frac{\sqrt{2}}{2}(-1+j)+\rho \exp(j\theta)}{(\frac{\sqrt{2}}{2}(-1+j)+\rho \exp(j\theta))^2+ \sqrt2(\frac{\sqrt{2}}{2}(-1+j)+\rho \exp(j\theta)))+1}}

The Attempt at a Solution

Solution is:
∞ exp( \frac{∏}{4} - \theta)

The denominator approaches 0 and the numerator doesn't. It doesn't have a limit.

 

[stefaneli]

To be exact...
\rho \rightarrow 0+

The solution I've written is correct for sure.:)

 

[Dick]

To be exact...
\rho \rightarrow 0+

The solution I've written is correct for sure.:)

Ok, let's write a=\frac{\sqrt{2}}{2} (-1+j) and r=\rho exp( j \theta) then your expression is \frac{a+r}{(a+r)^2+\sqrt{2} (a+r)+1}
If you expand the denominator, and putting in the value for a, you get \frac{a+r}{r^2+j \sqrt{2} r}
As ρ→0 you can ignore the r in the numerator and the r^2 in the denominator. Now you just have to express \frac{a}{j \sqrt{2} r} as a magnitude and phase. Can you take it from there? It's not really a limit, it's a limiting behavior.

 

[stefaneli]

Thanks...it helped me:)