Verified answer

In general, it is very difficult to work these out. But, typical textbook problems are rigged so that you can be succesful.

A handy trick is to realize that x^2 <= x^2 + y^2

Also, you need to convince yourself that a reasonable guess for the value of the limit is 0.

So you want | f(x,y) - 0| < eps

but | f(x,y)| <= | 2(x^2 + y^2) y/(x^2 + y^2)| = |2y|

this will be less than epsilon when |y| < eps/2

now |y| <= sqrt(x^2 + y^2) = | (x,y) - (0,0) |

so choose delta = eps/2.

When | (x,y) - (0,0) | < delta, that forces |y| < delta, which forces

| f(x,y) - 0| < eps

Delta Epsilon Proof Examples

assume e > 0. we want a d such that: 0 < sqrt((x - a million)^2 + (y - a million)^2) < d ===> |x^2 + y^2 - 2| < e working backwards: |x^2 + y^2 - 2| < e <=== |(x^2 - a million) + (y^2 - a million)| < e <=== |(x - a million)(x + a million) + (y - a million)(y + a million)| < e <=== |x - a million|.|x + a million| + |y - a million|.|y + a million| < e ... (by triangle inequality) If we anticipate sqrt((x - a million)^2 + (y - a million)^2) < a million, then: (x - a million)^2 + (y - a million)^2 < a million ===> (x - a million)^2 < a million and (y - a million)^2 < a million ===> |x - a million| < a million and |y - a million| < a million ===> |x - a million| + |2| < 3 and |y - a million| + |2| < 3 ===> |x - a million + 2| < 3 and |y - a million + 2| < 3 ... (triangle inequality back) ===> |x + a million| < 3 and |y + a million| < 3 So, below this assumption, we've that: |x - a million|.|x + a million| + |y - a million|.|y + a million| < e <=== 3|x - a million| + 3|y - a million| < e <=== |x - a million| + |y - a million| < e / 3 Now, if we stress sqrt((x - a million)^2 + (y - a million)^2) < e / 6, then |x - a million| and |y - a million| are below e / 6, and so: |x - a million| + |y - a million| < e/6 + e/6 = e/3 to that end if we make sqrt((x - a million)^2 + (y - a million)^2) below the two e/6 and a million, then we've shown the decrease. So, set: d = min{a million, e/6}