# Does $\lim_{(x,y)\to(0,0)}\frac{x^2+y^4}{x+y^2}$ exist, and equal $0$?

I just need a simple yes or no answer to see if my answer is in line with a consensus. Does the limit $$\lim_{(x,y)\to(0,0)}\frac{x^2+y^4}{x+y^2}$$ exist? I think it does, and the limit is 0.

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"I think it does, and the limit is $0$". Can you write down your reasoning? – Jack Apr 9 '12 at 4:49

Hint: Note the denominator is equal to zero along the parabola $x = y^2$. That's going to make it impossible for the limit to exist.
@shmiggens: The $\cos\theta$ in the denominator can be viewed as relevant, but be careful not to apply that reasoning to say $\dfrac{x^6+y^8}{x^2+y^4}$. – André Nicolas Apr 9 '12 at 5:50
Along the path $x=-y^2+y^5$, \begin{align*} \frac{x^2+y^4}{x+y^2} = \frac{2y^4+y^{10}-2y^5}{y^5}\rightarrow \infty. \end{align*} That is, \begin{align*} \lim_{(x, y)\rightarrow (0, 0)} \frac{x^2+y^4}{x+y^2} \end{align*} does not exist.