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I am reading on a proof of the product law for limits and one of the steps confuses me. I'll just show the important inequality and omit the detials

$|f(x)g(x) - f(a)g(a)| \leq |f(x)||g(x) - g(a)| + |g(a)||f(x) - f(a)| < |1 + f(a)|\dfrac{\epsilon}{2(|f(a)| + 1)} + |g(a)|\dfrac{\epsilon}{2(|g(a)|+1)} = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $

Where we assumed f and g have limits and that $|f(x) - f(a)| < \min \{1,\dfrac{\epsilon}{2(|g(a)| + 1)}\}$ and $|g(x) - g(a)|<\dfrac{\epsilon}{2(|f(a)|+1)}$

How on earth did they come up with those $\epsilon$? And how did $|g(a)|$ cancel out with $2(|g(a)|+1)$ in the bottom to give that 1/2?

I found this proof in Spivak's book. Is it a typo? Maybe it should be < instead of =?

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Yes, the $=\frac{\epsilon}{2}+\frac{\epsilon}{2}$ should be $\lt \dots$. And it wasn't cancellation, it was inequality. – André Nicolas Nov 12 '12 at 7:56
up vote 2 down vote accepted

We'd like to use $|f(x) - f(a)| < \dfrac{\epsilon}{ 2 |g(a)|}$ so that $|f(x) - f(a)| |g(a)| < \epsilon/2$ and $|g(x) - g(a)| < \dfrac{\epsilon}{2 |f(x)|}$ so that $|f(x)| |g(x) - g(a)| < \epsilon/2$, but this doesn't quite work.

First of all, $|g(a)|$ might be $0$, which would be bad in a denominator, so instead we use $|f(x) - f(a)| < \dfrac{\epsilon}{ 2 (|g(a)| + 1)}$. Then $$|f(x) - f(a)| |g(a)| < \dfrac{\epsilon |g(a)|}{2 (|g(a)| + 1)} < \dfrac{\epsilon}{2}$$

For the other one we have the additional problem that we need a fixed bound on the right, not one that depends on $x$. So we require additionally $|f(x) - f(a)| < 1$, which implies $|f(x)| < |f(a)| + 1$, and then if $|g(x) - g(a)| < \dfrac{\epsilon}{2 (|f(a)| + 1)}$ $$ |f(x)| |g(x) - g(a)| < (|f(a)| + 1) \dfrac{\epsilon}{2 (|f(a)| + 1)} = \dfrac{\epsilon}{2}$$

Adding the two inequalities up, we get

$$|f(x)| |g(x) - g(a)| + |f(x) - f(a)| |g(a)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $$

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Why do they add "min" for f's inequality? Can the term with epsilon evern even be greater than 1? – Hawk Nov 12 '12 at 8:34
As I said, we want both $|f(x) - f(a)| < \epsilon/(2(|g(a)|+1)$ and $|f(x) - f(a)| <1$. So to ensure that, we require $|f(x) - f(a)| < \min(1, \epsilon/(2(|g(a)|+1))$. – Robert Israel Nov 12 '12 at 9:11

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