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Arshjot S.

Calculus 3

4 months, 3 weeks ago

Show that $f(x, y)$ has no limit as $(x, y) \rightarrow(0,0)$ if $$f(x, y)=\frac{x y}{|x y|}, \quad x \neq 0 \text { and } y \neq 0.$$

Chapter 12

Functions of Several Variables

Section 6

Limits and Continuity

Yeah, we have a function F of X. Y. Ah defined here um noticed a restriction. Okay, so we have X. Y divided by the absolute value of X times life. And uh dysfunction is restricted. Uh It's not being defined when um X zero or Y equals zero. So here's the definition of the function F. Of X. Y. As long as X and Y are not zero. Well X equals zero on the entire Y axis. So we don't want to include the Y axis. So every point on the Y axis is not included in the domain of this function. Likewise wise not allowed to equal zero. Um At every point on the X axis, the Y coordinate would be zero. So since why is not allowed to equal zero? Uh The entire X axis is not part of the domain of this function. So here is our function F. Of X. Y. And it is defined on the entire xy plane. Except for the axis. We want to show that the limit of this function as we approach 2.0 which is here at the origin. We want to show that the limit of dysfunction as we approach zero comma zero does not exist. And here's how we are going to do that. First of all, let's study dysfunction. We have X times Y being divided by the absolute value of X times Y. So the X. Y and Z X. Y. They are the same. The only thing that's different is in the denominator you have an absolute value. So let me just give you a couple numbers real quick to give you an idea. Suppose X times Y is too. Then you would have to over the absolute value of two which would simply be well, absolute value to is to always gives you absolute value always gives you a positive number. Sotu over to would be what? But suppose x times Y came out to be negative three, then you would have negative three over the absolute value of negative three. In this case you would have negative three over positive three because absolute value turns everything positive and negative three over positive three would be negative one. So X Y divided by the absolute value of X. Y can be one or it can be negative one. It all depends on what X times Y or more specifically what sign X times Y. Hats, if x times Y is a positive a number then your function is going to equal one. If x times why comes out to be a negative number, uh then your function is gonna be negative one. Now in this first quadrant, in this first quadrant, uh your x coordinate is positive and your y coordinate is positive. So in this first quadrant um X times y will be positive. And that means that means our function f will equal one for every point in the first quadrant. Now let's take a look at the second quadrant in the second Clyde jane, your x coordinates are negative, why coordinates are positive? Now if you're x coordinate is negative and your y coordinate is positive, then X times y is going to be negative less than zero. And we just saw that if X times y is negative, your function, this function is going to come out to equal negative one. So for every point in this quadrant your function equals one for every point in this quadrant, your function has the value of negative one. So what happens um as we approach 2.0 if we approach 2.0 from a direction in the first quadrant, your function is going to stay fixed at one. So here your function is approaching the value of one as we approach 2.0 But if we approached, if we approach the uh if we approach 2.0 from a direction in the second quadrant like this, then uh you're function since its uh stays fixed at negative one in your second quadrant. As we approach 2.0 from a direction in the second quadrant, your function is going to approach approach the value of negative one. So the limit of the function as we approach 2.0 does not exist because if we approach the 0.0 from these two directions, the function is approaching two different values. And so the limit will not exist.

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