Questioning 3D Limits: What To Try Next?

  • Thread starter Thread starter Char. Limit
  • Start date Start date
  • Tags Tags
    3d Limits
Char. Limit
Gold Member
Messages
1,222
Reaction score
23
So, there was a big long post here with examples and such, but then my internet crashed. So screw the examples, I'll just get straight to my question.

When seeing if a limit exists in 3d (with the limit being (0,0)), I noticed that we often try the lines x=0 and y=0 first. Are these always the best lines to try, or are there others that seem intuitive? If both of these go to the same number, what should I try next?
 
Physics news on Phys.org
In general, trying y = mx is always nice. You pick whatever is easy, or whatever looks like it would cause a contradiction.

You can always try polar if it looks like it would work. If it works in polar, it covers all cases.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top