# How to expect if a limit exists before attempting squeeze theorem

From this video, I've learned to expect if a limit exists or not is by comparing the combined power on the numerator is bigger than than the combined power on the denominator.

For example, when considering limit as $(x,y) \rightarrow (0,0)$the rational function $$\frac{3x^2y}{x^2 + y^2}$$

The numerator has an exponential power of three ($x^2 * y)$ whereas the denominator has an exponential power of only 2.

However, I tried to use this intuition to surmise the existence of a limit as$(x,y) \rightarrow (0,0)$ for the rational function $\frac{x^4 - 4y^2}{x^2 + 2y^2}$.

As it has an exponential power of 4 on the numerator and an exponential power of 2 on the denominator, I expected the limit to exist. After struggling to prove this using the squeeze theorem, I reassessed using the path test, which quickly proved that it does not.

What are ways of quickly assessing whether we can expect a limit to exist?

• You need the lowest powered term in the numerator to be greater than the highest powered term in the denominator. The lowest powered term in $(x^4+4y^2)$ is $4y^2$ and too small to force the limit to converge at $0.$ – Doug M Sep 14 '17 at 22:02
• I see - so the rule of thumb is the lowest powered term in the numerator has to be greater than the highest powered term in the denominator ? Not comparing the overall exponential power of numerator vs denominator, as I had thought? – maddie Sep 14 '17 at 22:13
• also for the equation, $\frac{5y^4cos^2x}{x^4 + y^4}$ -- do we have to compare the lowest/highest power of the same term? In other words, do I compare $y^4$ on top vs $y^4$ on bottom - or can I say that because $cos^2$ is a smaller power than $y^4$, I can expect the limit to not exist? – maddie Sep 14 '17 at 22:52
• Don't get to wedded to simple rules. $cos^2 x$ is not going to $0$ as $(x,y),$ so you should just disregard that factor, and treat it like it is a constant. And as you observed in the original post if you have $\frac {x^ny^m}{x^a+ y^b}$ you can consider the power of the numerator to be $(m+n)$ but if you have $\frac {x^n+y^m}{x^a+ y^b}$ then the power of the numerator (for this calculation) is $\min(m,n)$ – Doug M Sep 14 '17 at 23:03
• And if the denominator is $x^a + y^b$ then the power of the denominator is $max(a, b)$ ? – maddie Sep 14 '17 at 23:49

Using polar coordinates $$x=r\cos (t) \;\;\; y=r\sin (t)$$
$$f (x,y)=g (r,t)=3r\sin (t)\cos (t)$$
as $r\to 0$, $g (r,t)$ goes to zero for all $t$. So the limit exists and is zero.