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Hi I have problem to decide if
enter image description here

is continuous at (0,0).
I have tried to slove this by polar cordinates
enter image description here

but from here the only thing I can tell is , that for every teta, the limit will be 0.

Is it Enough? because it seems to me that in this case I am only showing that for every linear path to (0,0) the limit will be 0 ? and is this function really continuous?

one more thing I heard about a Rule of thumb that if the sum of the powers in the numerator is lower then the denominator the limit doesn't exist. is it allways true? thanks,

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The rule of thumb would appear to be false for $x/(x^2+1)$. – Gerry Myerson Aug 23 '13 at 13:04
I think the rule of thumb that you have heard is supposed to be the reverse: If the sum of powers in the denominator is less than the sum of powers in the numerator, the limit doesn't exist... But I'm not sure how that transfers to multivariate calculus. – apnorton Aug 23 '13 at 13:14
@anorton, $x^2/(x+1)$ would appear to be a counterexample to your statement. – Gerry Myerson Aug 24 '13 at 7:03
@GerryMyerson Oh. I had saw the "rule of thumb" the OP wrote out, and my mind jumped to asymptotics... if we were only talking about limits $\to \pm \infty$, then my statement would be correct, methinks. – apnorton Aug 24 '13 at 12:52
up vote 1 down vote accepted

Hint: Note that $\dfrac{y^2}{x^4+y^2}\le 1$.

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Go by the definition. Fix $\epsilon>0$. Let $\delta=\epsilon^{2/3}$, ans suppose that $\sqrt{x^2+y^2} \leq \delta$. You need to prove that $|\frac{x^{3/2}y^2}{x^4+y^2}| \leq \epsilon$. Indeed, by the assumption we have $|x| \leq \delta = \epsilon^{2/3}$, and so $$ |\frac{|x^{3/2}|y^2}{x^4+y^2}| \leq |x^{3/2}| \cdot \frac{y^2}{x^4+y^2} \leq |x^{3/2}| \leq \epsilon, $$ as required.

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Ok, thanks for the help? but just to verify my way to slove this question is good enough? with the polar cordinates ? – tAmir Aug 23 '13 at 13:30
@user, showing it works for every linear path is not good enough --- something different could happen along a curve. But the polar coordinate method works, if you show that even for variable $\theta$ the limit is zero. – Gerry Myerson Aug 24 '13 at 7:08

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