# Homework Help: Limits of 2 var functions approaching paths of parabolas

1. Aug 26, 2011

### davidp92

1. The problem statement, all variables and given/known data
http://i.imgur.com/F6qLL.jpg
In the example, it says if the approaching path of the limit is y=x^2 and x=y^2, the limits are 0. I can see the limit is 0 for approaching in the path of x=y^2 since the function becomes f(x,y) becomes f(y^2,y)=3y^5/(y^4+y^2) $\rightarrow$ 0 as (y^2,y) $\rightarrow$ 0.
But doing the same for the path of y=x^2, I get f(x,x^2)=3x^2*x^2/(x^2+x^4)=3x^4/(x^2+x^4) which $\rightarrow$ 3 as (x,x^2) $\rightarrow$ 0. What am I doing wrong?

2. Aug 26, 2011

### LCKurtz

$$\frac {3x^4}{x^2+x^4}=\frac{3x^2}{1+x^2}\rightarrow \frac 0 1$$

3. Aug 26, 2011

### SammyS

Staff Emeritus
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4. Aug 27, 2011

### HallsofIvy

Notice that your example says "we begin to suspect that the limit exists and is 0." (emphasis mine)

Frankly, I don't know what they are trying to show. The fact that the limit is 0 along some curves may suggest that the limit itself is 0 but it certainly doesn't have to be. I don't see that doing 4 paths rather than 2 tells us anything at all.

If you convert to polar coordinates, the function becomes
$$\frac{3r^3 cos^2(\theta)sin(\theta)}{r^2}= 3r cos^2(\theta)sin(\theta)$$
which goes to 0 as r goes to 0 no matter what $\theta$ is. That tells us that
$$\lim_{(x,y)\to (0, 0} \frac{3x^2y}{x^2+ y^2}= 0$$

Because distance from the point to the origin is given by the single variable, r, as long as the limit, as r goes to 0, is independent of [itex]\theta[/tex], then that limit exists.