Recent content by Drako Amorim

It's a disk with radius approaching to 0. In case 0 <= t <= 2pi is independent and the limit become: \lim_{r\rightarrow 0}\frac{r^3 \cos^3(t)+r^3 \sin^3(t)}{r\cos(t)+r^2\sin(t)}. If t=pi/2 or 3pi/2 -> cos(t)=0 and sin(t)=+-1 and the limit converge to 0. Else we can do this: \lim_{r\rightarrow...

Yes. x=t and y=0 so the limit will be: \lim_{(t)\rightarrow (0)}\frac{\sin t^{3}}{t}=\lim_{(t)\rightarrow (0)}t^2\frac{\sin t^{3}}{t^3}=0. x=0, y=t: \lim_{(t)\rightarrow (0)}\frac{\sin t^{3}}{t^2}=\lim_{(t)\rightarrow (0)}t\frac{\sin t^{3}}{t^3}=0. x=y=t: \lim_{(t)\rightarrow (0)}\frac{\sin...

ok. L'hospital rule was applied in this limit: \lim_{(0,y)\rightarrow (0,0)}\frac{\sin y^{3}}{y^{2}}. edit: In fact this was a dumb way to solve this simple limit. You have a proof for this? Because paths do not are sufficient. Can you show me in algebra what you see?

The limit is not 0? If you apply the L'hospital method it will give 0. Can you show me what you're saying? You are saying that you can separate \frac{sin(x^3+y^3)}{x+y^2} into f(x)g(y)? About your question: This limit is a challenge that came from \lim_{(x,y)\rightarrow (0,0)}\frac{\sin...

Can you be more clear?

\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x^{3}+y^{3}}\frac{x^{3}+y^{3}}{x+y^{2}}=\lim_{(x,y)\rightarrow (0,0)} \frac{x^{3}+y^{3}}{x+y^2}=\lim_{(x,y)\rightarrow (0,0)} \frac{(x+y^{2}-y^{2})x^{2}+(x+y^{2}-x)y}{x+y^2}=\lim_{(x,y)\rightarrow (0,0)}...

Ok. \lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x^{3}+y^{3}}\frac{x^{3}+y^{3})}{x+y^{2}}=\lim_{(x,y)\rightarrow (0,0)}\frac{x^{3}+y^{3}}{x+y^{2}} The only progress that I have was reduce the limit to this: \lim_{(x,y)\rightarrow (0,0)}\frac{\sin...

I'm trying to verify that: \lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x+y^{2}}=0. 0<\sqrt{x^2+y^2}<\delta\rightarrow |\frac{\sin (x^{3}+y^{3})}{x+y^{2}}|<\epsilon 0\leq |\sin (x^{3}+y^{3})|\leq |(x^{3}+y^{3})|\leq |x|x^2+|y|y^2 |\frac{\sin (x^{3}+y^{3})}{x+y^{2}}|\leq \frac{...

Redraw the Delta circuit for a Pi And I conclude that: Vac = (q1 - q3)/Cy Vbc = (q2 + q3)/Cx Vab = q3/Cz = Vac + Vcb = Vac - Vbc → q3/Cz = (q1 - q3)/Cy - (q2 + q3)/Cx → q3 (Cy Cz + Cx Cz + Cx Cy)/(Cx Cy Cz) = q1/Cy - q2/Cx ∴q3 = Cx Cy Cz / (CyCz + CxCz + Cx Cy) [q1/Cy - q2/Cx] ≡ K...

This problem came from the twelfth edition Sears & Zemansky Electromagnetism, is one of the challenging problems. I think that I solved it and I will transcribe it, but I think that problem is unfair, because there's no 2-port network in the main text

Homework Statement Hello everyone. The problem is related to these two capacitor connections: In the enunciate it is said that these two are equivalent and it is requested to prove that: C1=(CxCy+CyCz+CzCx)/Cx C2=(CxCy+CyCz+CzCx)/Cy C1=(CxCy+CyCz+CzCx)/Cz Homework Equations C=Q/Vab...

Sorry, I have posted in the wrong place :'(

Hi everyone, the problem is related to these two capacitor connections: In the enunciate it is said that these two are equivalent and it is asked to prove that: My problem with these solution is that I do not know how analyze them, because of the two potencial difference. Can you please...