$$f(x,y)=\frac{2(x^3+y^3)}{x^2+2y}$$
Show that $f$ is not continuous at $(0,0)$ when it is given that $f(x,y)=0$ at $(0,0)$.
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Normally substituting $y=mx^n$ works.. not in this case – Prayagdeep Parija Jan 16 '14 at 13:06
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what do you mean when you say "not in this case" – Jan 16 '14 at 13:07
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in here it is not working – Prayagdeep Parija Jan 16 '14 at 13:08
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substituting it doesn't reduce it to the form \frac{1+m^p}{A+m^q} which would have done the job – Prayagdeep Parija Jan 16 '14 at 13:08
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Ok... /i have just now checked.. They does not help much ..how about taking polar coordinates?? $x=\cos \theta$ and $y=\sin \theta$? – Jan 16 '14 at 13:11
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What is $$f\left(x,-\frac{x^2}{2}\right)$$ for $x\neq 0$? – Daniel Fischer Jan 16 '14 at 13:30
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@ praphulla .. I am not sure how $cos\theta$, sin$\theta$ helps?? – Prayagdeep Parija Jan 16 '14 at 13:56
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@ daniel .. Not defined at $(x,-\frac{x^2}{2})$.. Can i get something like $y=mx^n$ where the whole limit will depend upon $m$?? – Prayagdeep Parija Jan 16 '14 at 13:58
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@Prayagdeep My point is that the denominator vanishes there, but the numerator doesn't. So if you approach the origin along a curve sufficiently close to that, you can obtain the limits $\pm \infty$. – Daniel Fischer Jan 16 '14 at 14:07
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@ daniel .. thank you.. got it – Prayagdeep Parija Jan 16 '14 at 14:10