Inequality proof with absolute values

Question

How do you prove the following:

$$ \varepsilon > 0 \\ \left | y-b \right | < \varepsilon\\ \left | x-a \right | < \varepsilon \\ \Longrightarrow \left | xy - ab \right | < \varepsilon (\left | a \right | + \left | b \right | + \varepsilon)$$

Thanks in advanced

Answer 1

$$|xy-ab|=|xy -bx + bx - ab|=|x(y-b)+b(x-a)|\leq|x|\cdot|y-b|+|b|\cdot|x-a|$$ This last inequality comes from the triangle inequality $|a + b| \leq |a| + |b|$.

Now since $|y-b|<\varepsilon$ and $|x-a|<\varepsilon$ we have $|x|\cdot|y-b|+|b|\cdot|x-a| < |x|\cdot \varepsilon + |b|\cdot \varepsilon$. More concisely stated, we now have $$|xy-ab| < \varepsilon(|x| + |b|). \tag{1}$$

Now, using the "alternate triangle inequality" $|a-b| \geq |a| - |b|$, note that $|x-a| < \varepsilon$ therefore implies that $|x|-|a|<\varepsilon$, or $$|x|<\varepsilon + |a|. \tag{2}$$

Combining $(1)$ and $(2)$, we get that $$|xy-ab|<\varepsilon(|x|+|b|)<\varepsilon(|a|+|b|+\varepsilon).$$

Answer 2

$\begin{align}|xy-ab|&=|xy+ya-ya+xb-xb+ab-ab-ab| \\ &=|b(x-a)+a(y-b)+y(x-a)+b(a-x)|\\&<|b(x-a)|+|a(y-b)|+|y(x-a)+b(a-x)|\\& <|b(x-a)|+|a(y-b)|+|(x-a)(y-b)|\\&<|b||(x-a)|+|a||(y-b)|+|(x-a)||(y-b)|\\&<|b|\epsilon + |a| \epsilon + \epsilon \epsilon\\&=\epsilon(|a|+|b|+\epsilon) \end{align}$