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By considering different- paths of approach, show that the function below hlx,y) = has no limit as (xy)--(0,0)Examine the values of h along curves thatend at (0,0)...

Question

By considering different- paths of approach, show that the function below hlx,y) = has no limit as (xy)--(0,0)Examine the values of h along curves thatend at (0,0) Along which set of curves is h OA y=kx? constant value? X#0,k#0 0 B. y=kx + kx?,xFO,kf0 C. y=kx,X#0,k#0 0 D. y=k,XFO,kf0 If (X,y) approaches (0,0) along the curve when k = _ used in the set of curves found above; what is the limit? (Simplify your answer:) If (X,y) approaches (0,0) along the curve when k =2 used in the set of curves f

By considering different- paths of approach, show that the function below hlx,y) = has no limit as (xy)--(0,0) Examine the values of h along curves thatend at (0,0) Along which set of curves is h OA y=kx? constant value? X#0,k#0 0 B. y=kx + kx?,xFO,kf0 C. y=kx,X#0,k#0 0 D. y=k,XFO,kf0 If (X,y) approaches (0,0) along the curve when k = _ used in the set of curves found above; what is the limit? (Simplify your answer:) If (X,y) approaches (0,0) along the curve when k =2 used in the set of curves found above, what is the Iimit? (Simplify your answer:) What can you conclude? Since f has the same limit along two different paths to (0,0), by the two-F ~path test; has no limit as (x,Y) approaches (0,0}. has two different limits along two different paths to (0,0}, by the two-path test; f has no limit as (X,Y) approaches (0,01. Since different paths to (0,0), it cannot be determined whether or not / has limit as (x,y) approaches (0,0}. Since has the same limit along two has limit as (X,Y) approaches (0,0} paths to (0,0), it cannot be determined whether or not = Since has two different limits along two different



Answers

Let $$f(x, y)=\left\{\begin{array}{cl}\frac{x y^{3}}{x^{2}+y^{2}} & \text { if }(x, y) \neq(0,0) \\ 0 & \text { if }(x, y)=(0,0)\end{array}\right.$$,(a) Compute the limit as $(x, y) \rightarrow(0,0)$ of $f$ along the path $x=0$ (b) Compute the limit as $(x, y) \rightarrow(0,0)$ of $f$ along the path $x=y^{3}$ (c) Show that $f$ is not continuous at (0,0)

Yeah, it's in this case, we will consider two different patents to take our limit of the given function. S so, in case we can consider the Y axis, that is with X zero. Who will be approaching your origin on my access? Uh, this implies you image. Ah, zero squared. Plus why, bye bye. Why, which is one? And alternatively, we could consider approach alone the curve exit route. Why? Which of course, is the same as y x squared, the approachable, proble, the origin. And in this case route, why squared? That's why way will divide by why? And so, of course, the seal of the route. Why square turns Why plus y, which is to why over why, which is, uh too. And so we're calling that the limit exists only independently of the curve on wish it is approached. We find that it does not exist, which is what we set out to do because approaching it along two different parents. One case Why accidents? Another the problem of equation wise X squared. We find two different values respectively. One in two

To find the solution off this problem, we will assume why equals constant K. I want to play our body will x square. So we would have makes a square plus key excess square over key excess square. No way can remove or a common factor X square. So our solution will be equals one plus k musky over p which represent different limits for different values of key where key cannot be equal. Cereal. Thank you.

This question I have the h of X y is equal to x squared times Why over excellent forth plus y squared ona show This lemon does not exist And I want to do that by showing that if you go along a different pathway, your limit is goingto be a different value. So for this one, I'm going to go along. Why is equal tio K X squared? And the reason I wouldn't do that is because I have this extra forthe right here. I kind of want these to combine together somehow. This isn't squaring this Someone take a X squared so that means I won't have the limit is X why approaches 00 the function And so that's going to equal to limit as X approaches zero as X squared. Okay, X squared over x 1/4 plus case. Grid square, which is going to equal to the top, is just going to be k times x the fourth. This is really case grune times extra force so I can factor out next to the fourth and I'm left with K squared plus one next to the fourth and some extra force cancel and I'm left with K over one plus case squared. And so, looking at this for all different values of K, I'm going in a different woman, and so that shows that the limit does not exist.

To find the solution off this level. We less you, my equals constant K. I want to play our valuable extra square. So you get export four Kate over export for class X. Our floor. Thanks. Or four King or talked. So we now we can't substitute. Or so you remove Simplify export for as a common factor before our final solution e k over one plus que square which represent different values. Off limit for different values. A key? Um, yeah. Thank you.


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