# 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.

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### 1 Answer

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)=`

`lim_(r->0)(r^5cos^3thetasin^2theta)/(r^4)=lim_(r->0)r(cos^3thetasin^2theta)=0,`

` ` 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

`lim_{(x,y)->(0,0)}(x^2y^2)/((x^2+y^2)^2)=lim_(r->0)(r^4cos^2thetasin^2theta)/(r^4)=`

`lim_(r->0)(cos^2thetasin^2theta).`

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.

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