I'm getting 2 different values for this limit but am i doing it right/

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Hello everyone, I'm having troubles deciding if this limit does not exist...
i have the problem and work here, i get 2 different limits by letting x and y go to different points and making z fixed at 0. Did i break any rules?

http://img426.imageshack.us/img426/1372/lastscan7fk.jpg
 
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You forgot to indicate your parameterization in the second attempt. What was it?
 
It seems to have been (t,t,0).
 
correct (t,t,0)
 
Yes, in order for the limit to exist, f(x,y,z) must be close to that limit when (x,y,z) are close to 0, no matter how you approach (0,0,0). Since you get two different limits approaching along two different lines, this limit does not exist.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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...
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