cristobalsele

1. Comunidad
2. Study Group
3. evaluate limit

Since 1<=sin^2(2pi/x)+1<=2 for all x and since lim x->0 sqrt(x) = 0 you will have that limx->0 sqrt(x)*(sin^2(2pi/x)) = 0   Note: Have in mind that limx->+0 2pi/x = infinity, so sin^2(infinity) wil...

1. Comunidad
2. Study Group
3. Angle of complex numbers

Maybe this can help you. z=x+yi z=-5+3i -> x<0 & y>=0 so arctan(y/x)+pi to  

1. Comunidad
2. Study Group
3. evaluate limit

So normally a function for example like f(x)=x or f(x)=x^2 and evaluate it's limit it tends to go into a specific value, in bot examples it would be lim x->inf = inf. However when you evaluate func...