MthT 430 Spivak Problem 1.21

21.

Assume that if

|x - x0| < min æ è e 2(|y0| + 1) ,1 ö ø , |y - y0| < e 2(|x0| + 1) .

then

|x y - x0 y0| < e.

The trick is to write x y - x₀ y₀ in terms of x -x₀ and y - y₀. There are several ways to do this, but the one which works best is

x y - x0 y0 = x(y - y0 + y0) - x0 y0 = x (y - y0) + (x - x0) y0 = I + II.

Then

|I| £ |x||y - y0| £ (|x0| + |x - x0|)|y - y0| £ (|x0| + 1)|y - y0| < e/2, |II| £ |x- x0||y0| < e/2.

Thus

|x y - x0 y0| £ |I| + |II| < e/2 + e/2 = e.

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On 04 Aug 2007, 15:37.