24a9f1…: “Up to a great castle”

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This exercise will establish an important class of function that can be used to show that a function may tend to the same value along all paths of a certain type to a point, but for which the limit does not exist at that point. Consider the function $F(x,y) = \frac {2x^2y}{3x^4+4y^2}$ .

Let’s try to analyze the function along all straight line paths through $(0,0)$ . Note that the most general form of a straight line path to $(0,0)$ is found from the point-slope form of a line below.

$y-y_0=m(x-x_0)$ Since the line passes through $(0,0)$ , we find that any straight line path through $(0,0)$ must be of the form $y=mx$ .

Now, let’s see what $F(x,y)$ approaches as $\vector {x,y} \to \vector {0,0}$ along $y=mx$ .

$F(x,y) = F(x,mx) = \frac {2x^2\left (\answer {mx}\right )}{3x^2+4\left (\answer {mx}\right )^2} = \frac {\answer {2m} \cdot x^3 }{3x^4 + \answer {4m^2}\cdot x^2} = \frac {2mx}{3x^2+4m^2}, x \neq 0.$

For any finite $m$ , note that as $x \to 0$ , $F(x,y) \to \answer {0}$ as $\vector {x,y} \to \vector {0,0}$ along $y=mx$ . If we approach $(0,0)$ along the $y$ -axis, we also find $F(x,y) \to \answer {0}$ either by taking $m \to \infty$ or by evaluating $F(x,0)$ for $y \neq 0$ .

Hence, along any straight line path, $F(x,y) \to \answer {0}$ as $\vector {x,y} \to \vector {0,0}$ .

Is this enough to determine that $\Lim {x,y}{0,0} = 0$ ? YesNo

We really must establish that $F(x,y)$ approaches the same value along any path, and there are other paths along which $(x,y)$ can approach $(0,0)$ .

Let’s consider the path $y=x^2$ .

First, can $\vector {x,y} \to \vector {0,0}$ along this path? YesNo

Now, note that along this path

$F(x,y) = F(x,x^2) = \frac {2x^2\left (\answer {x^2}\right )}{3x^4+4\left (\answer {x^2}\right )^2} = \frac {2x^4}{\answer {7} \cdot x^4} = \frac {2}{7} , x \neq 0.$

Hence, $F(x,y) \to \answer {\frac {2}{7}}$ as $\vector {x,y} \to \vector {0,0}$ along the path $y=x^2$ .

Can we determine now whether $\Lim {x,y}{0,0}$ exists?

Yes; $\Lim {x,y}{0,0} F(x,y)$ exists. Yes; $\Lim {x,y}{0,0} F(x,y)$ does not exist. No

This is an example of a function whose domain is all of $\R ^2$ except the origin, which allows us to consider any path that approaches $(0,0)$ in the domain.

This example is also not hard to modify to show that even if a function tends to the same value as $\vector {x,y} \to \vector {0,0}$ along all quadratic paths $y=a_2x^2+a_1x$ , this is still not sufficient to show that $\Lim {x,y}{0,0}$ exists. An example of such a function is

$F_3(x,y) = \frac {x^3y}{x^6+y^3}.$

This will tend to $0$ as $\vector {x,y} \to \vector {0,0}$ along $y=a_2x^2+a_1x$ for any choice of $a_1$ and $a_2$ , but will approach $1/2$ as $\vector {x,y} \to \vector {0,0}$ along $y=x^3$ .

It’s not hard to continue playing this game; the function

$F_4(x,y) = \frac {x^4y}{x^8+y^4}$

will tend to $0$ as $\vector {x,y} \to \vector {0,0}$ along $y=a_3x^3+a_2x^2+a_1x$ for any choice of $a_1$ , $a_2$ , and $a_1$ , but will approach $1/2$ as $\vector {x,y} \to \vector {0,0}$ along $y=x^4$ , and so on.

This same example can be modified to construct an example of a function that tends to the same value along any path for which $x^2+y^2 \to 0$ at any rate slower than a prescribed one, but for which the limit does not exist.

The lesson?

When we say that the function must tend to the same value along every path, we mean it!