# Determine the largest subset of the domain where the function is continuoush(x,y) = {`(2x^2+y^2)/(sqrt(x^2+y^2))` if (x,y) ≠ (0,0) and {0 if (x,y) = (0,0).

pramodpandey | Student

A function h(x, y) is continuous at the point (a, b) if the following two conditions are satisfied

(a)`h(0,0)` exists;

(b) `lim_{(x,y)->(0,0)}h(x,y)=h(0,0)`

h(0,0)=0   it is given.

Let `y=mx`  ,m is any real number.

`lim_(x->0)y=lim_(x->0)(mx)=0`

`h(x,y)=(2x^2+m^2x^2)/sqrt(x^2+m^2x^2)`

`=(x(2+m^2))/sqrt(1+m^2)`

`lim_{(x,y)->(0,0)}h(x,y)=lim_(x->0)(x(2+m^2))/sqrt(1+m^2)=0`

`Thus`

`lim_{(x,y)->(0,0)}h(x,y}=h(0,0)`

`So` h(x,y) is continous at every point of RxR.

Thus  largest subset is RxR.

oldnick | Student

If  `(x;y)!=(0;0)`  the function is contiune for  every `(x;y) in RR xx RR`  cause ratio of two continues function.

Concerning  `(x;y)=(0;0)`  we have to see what happens reaching the  point `(0;0)`  by means different direction.

Let it be  `t stackrel(->)(v)`  a direction.

Let run along side line:

`x=y`

Then  `x=t stackrel(->)(v)`   `y= t stackrel(->)(v)`

So we can approcah to  `(0;0) `   by means t goes close to zero.

`lim_(t->0) (2x^2+y^2)/sqrt(x^2+y^2)=` `lim_(t->0) (2t^2v^2+t^2v^2)/sqrt(t^2v^2+t^2v^2)=` `lim_(t->0) (3t^2v^2)/sqrt(2t^2v^2)=`

`lim_(t->0) (3t^2v^2)/(tv sqrt(2))=` `lim_(t->0) (3tv)/sqrt(2)=0`

It looks like  exits limit in `(0;0)`  so the function has a discontinuity in the origin eliminable.

To make it sure have to prove limit actually exists.

To reach this goal, have to see, wahtever line we approach to origin (even tough not stragiht line) limit is the same.

indeed le 's use following parametric function:

`x=x(t)` ;`y=y(t)` , function has the feature:

`lim_(t->0) x(t)= lim_(t->0) y(t)=0`

Since related by a link the two function ,approaching to zero has to be a finite ratio, then:

`lim_(t->0) (x(t))/(y(t))= k`

calculating limit:   `lim_(t->0) (2x^2(t)+y^2(t))/sqrt(x^2(t)+y^2(t))=`

`=lim_(t->0) ((2k^2+1)y^2(t))/sqrt((k^2+1)y^2(t))` `=lim_(t->0) y(t)(2k^2+1)/sqrt(k^2+1)=0`

As we'r awaiting

Then function is continue on `RR xx RR`  with a eliminable  discontinuity in the origin.