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1. ### Limit of COMPLEX-valued functions

2a) Being close to a complex number is the same as being close to a real number. Your function will approach that value. I'm not exactly sure where your confusion is so it is kind of hard to answer this. 2b) If your function allows i to be factored out then you can do that. It can be treated...
2. ### Changing from rectangular coordinate to sperical

Look at the third to the last line again: cos^2(phi) = sin^2(phi)[cos^2(theta) + sin^2(theta)] hope that helps.
3. ### Limit of COMPLEX-valued functions

1) If you are trying to visualize what it looks like in a real plane, then the function would only exist when (wx) is n*pi (n=0,1,2...). You would have a discontinuous function. If you consider the complex plane (where the vertical axis is imaginary and the horizontal axis is real), you could...
4. ### Limit of COMPLEX-valued functions

Math doesn't really change when you deal with complex numbers. You can ignore the "i" if you just want to evaluate a limit. Hope that helps.
5. ### Limit of COMPLEX-valued functions

Sine and cosine are bounded because they always return values between -1 and 1. If you have an imaginary component, that doesn't change the value sine will return. If you graph the function on a complex plane it will oscillate as x goes to infinity. It will not blow up or converge.
6. ### Rectangle to Cylindrical coordinate question

Those are not x, y, and z limits anymore once you convert to cylindrical coordinates. They are limits of theta, r, and z. There is not an x to integrate.
7. ### Limit of COMPLEX-valued functions

exp(-3x) goes to zero because exp(3x) is in the denominator. As for the cos(wx)-isin(wx), it will just oscillate back and forth as x goes to infinity. So you are multiplying zero and an oscillating function.
8. ### Rectangle to Cylindrical coordinate question

It may help if you start by sketching what the figure looks like. Then from the boundaries you can express it as r, theta, and z. From that you can find your r*d(theta), dr, and dz.
9. ### Implicitly differentiating PDE (multivariable calculus)

That's right because you have the 3z2x(dz/dx) + z3(dx/dx). I don't know what I was thinking... Thank You!
10. ### Implicitly differentiating PDE (multivariable calculus)

The problem: Find the value of dz/dx at the point (1,1,1) if the equation xy+z3x-2yz=0 defines z as a function of the two independent variables x and y and the partial derivative exists. I don't know how to approach the z3x part. I thought you would use the product rule and get 3(dz/dx)2x +...