a) Find  `lim_{(x,y)->(0,0)} [(x^3y^2)/((x^2+y^2)^2)]` b) Find `lim_{(x,y)->(0,0)} [(x^2y^2)/((x^2+y^2)^2)]`                 Polar coordinates may be used.

1 Answer | Add Yours

degeneratecircle's profile pic

degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted on

For both problems, we use the relationship between cartesian and polar coordinates. If a point has cartesian coordinates `(x,y)` ` ` and polar coordinates `(r,theta),` then

`x=r costheta` and `y=rsintheta.` This means that `x^2+y^2=r^2(cos^2theta+sin^2theta)=r^2.`

Also, `(x,y)->(0,0)` becomes `r->0.` Making these substitutions we see that

a) `lim_{(x,y) ->(0,0)} (x^3y^2)/((x^2+y^2)^2)=lim_(r->0)((rcostheta)^3(rsintheta)^2)/((r^2)^2)=`


` ` since `costheta` and `sintheta` are bounded and `r` going to `0` forces the whole expression to go to `0` . The limit is 0.

b) This is very similar in terms of how we solve it, but the answer is different. We make the same substitutions as above to get



This limit is independent of `r,` and it will be different depending on which direction we approach `(0,0)` from (changing `theta` changes this direction). Thus the limit does not exist.

There is another way to see that it depends on direction: suppose we approach the point `(0,0)` along the line `y=x.` In that case, since `x` and `y` are equal, we always have

`(x^2y^2)/((x^2+y^2)^2)=x^4/((2x^2)^2)=x^4/(4x^4)=1/4,` so the limit in this direction is `1/4.`

However, if we approach along the x axis, so that `y=0,` then we have

`(x^2y^2)/((x^2+y^2)^2)=0,` so the limit in this direction is `0.` If the limit exists, it can't depend on which direction we approach from, so we have again shown that the limit doesn't exist.


We’ve answered 319,644 questions. We can answer yours, too.

Ask a question