Applying the definition of a limit

[sinbad30]

Homework Statement

Applying the definition of a limit to show that

lim ((x^3 * y(y-1) ) / (x^2 + (y-1)^2) = 0 as (x,y) approaches (0,1)

The Attempt at a Solution

|x| = sqrt(x^2) <= sqrt((x^2 + (y-1)^2))
|y-1|=sqrt((y-1)^2)<= sqrt((x^2 + (y-1)^2))
|y|<= |y-1| + 1 via the triangle inequality

Let e>0. We want to find d>0 such that

0 < sqrt((x^2 + (y-1)^2)) < d then (|x|^3 |y| |y-1|)/ (x^2 + (y-1)^2) < e

So

(|x|^3 |y| |y-1|)/ (x^2 + (y-1)^2) <=
[(sqrt((x^2 + (y-1)^2)))^3 * sqrt((x^2 + (y-1)^2)) * (sqrt((x^2 + (y-1)^2)) + 1) ] / (x^2 + (y-1)^2)

Any ideas as to how to break down the right side of the inequality?

 

[vela]

Homework Statement

Applying the definition of a limit to show that

lim ((x^3 * y(y-1) ) / (x^2 + (y-1)^2) = 0 as (x,y) approaches (0,1)

The Attempt at a Solution

|x| = sqrt(x^2) <= sqrt((x^2 + (y-1)^2))
|y-1|=sqrt((y-1)^2)<= sqrt((x^2 + (y-1)^2))
|y|<= |y-1| + 1 via the triangle inequality

Let e>0. We want to find d>0 such that

0 < sqrt((x^2 + (y-1)^2)) < d then (|x|^3 |y| |y-1|)/ (x^2 + (y-1)^2) < e

So

(|x|^3 |y| |y-1|)/ (x^2 + (y-1)^2) <=
[(sqrt((x^2 + (y-1)^2)))^3 * sqrt((x^2 + (y-1)^2)) * (sqrt((x^2 + (y-1)^2)) + 1) ] / (x^2 + (y-1)^2)

Any ideas as to how to break down the right side of the inequality?

It's helpful to recognize that $r = \sqrt{x^2 + (y-1)^2}$ is the distance of the point (x,y) from the point (0,1). In terms of r, the righthand side is $\frac{r^4(r+1)}{r^2} = r^2(r+1)$. Can you take it from there?