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If it does exist, do a $\delta$ and $\epsilon$ proof to show the limit really is as you claim. Show at least 4 cases when determining potential limits.

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A first easy step is to test radial limits:


$$\lim_{x\to 0}\frac{m^3x^5}{(2+m^2)x^2}=\lim_{x\to 0}\frac{m^3}{2+m^2}x^3=0$$

This strongly suggests (though doesn't prove) that the limit exists and is zero. If you need $4$ cases just pick $4$ unique values of $m$. The reason I would really expect this is because the numerator is of higher degree than the denominator, and so should drive values down sufficiently fast to overpower the tendency of the denominator to make the expression diverge. In this case I would express the problem in polar coordinates:

$$\lim_{r\to 0}\frac{r^5\cos^2\theta\sin^3\theta}{r^2(2\cos^2\theta+\sin^2\theta)}=\lim_{r\to 0}\ \ r^3\frac{\cos^2\theta\sin^3\theta}{\cos^2+1}$$



So if someone gives demands an error of $\epsilon$ you can say to take the disk of radius $\delta=\epsilon^{1/3}$.

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can you provide a reference for "radial limits"? – Vicfred Oct 25 '12 at 4:13
@Vicfred reference in what sense? A radial limit is just one where you approach the limit point along a straight line (i.e, radially). Setting $y=mx$ handles all of these radial lines at once except for $x=0$, which is trivial to check. – Robert Mastragostino Oct 25 '12 at 5:22

If you don't like polar coordinates, you could also note that $x^2 \leq 2x^2 + y^2$ and so $\frac{x^2}{2x^2 + y^2} \leq 1$. Hence $$\bigg| \frac{x^2y^3}{2x^2 + y^2} \bigg | = |y|^3 \frac{x^2}{2x^2 + y^2} \leq |y|^3$$ and since $|y| \leq \sqrt{x^2 + y^2}$, given $\epsilon > 0$, set $\delta = \epsilon^\frac{1}{3}$. Then $\sqrt{x^2 + y^2} < \delta$, implies $|y|^3 < \epsilon$, and so this bounds the whole thing by $\epsilon$.

With these sorts of weird ratio problems, you usually want to pick out the largest power in the denominator and try to control that with a power in the top, hoping there's enough left over to send the ratio to zero. For instance, we had a whole $|y|^3$ left in the top for this problem.

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