Function, using two-path test, finding limits

gap0063
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f(x,y)=(x)/ (1-y2), y does not equal 1 or -1

Is it possible to define a new function g(x,y) that is defined and continuous for all (x,y) in R2, and such that g(x,y)=f(x,y) for all (x,y) in the domain of f? If so, find such a function. If not, explain why.


I really don't know where to begin...
 
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For the existence of such g it is necesarry that lim (x->x0, y->1) f(x,y)=A(x0,1) exist and depend only on x0 (and not on the path). But in our case it does not hold:

Calculate the limit for x0=0 for two different paths:
1. y=x+1 (x->0)
2. y=2x+1 (x->0)

You won't get the same result, so such g does NOT exist.
 
how do i choose the equations of lines c and d,to find limits along when finding limits by two parts
 
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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