Theorom 6.1
- If $f$ and $g$ are continuous at $a$, then
- (1) $f + g$ is continuous at $a$
(2) $f \cdot g$ is continuous at $a$- Moreover, if $g(a) \neq 0$, then
- (3) $1/g$ is continuous at $a$
Theorom 6.3
Suppose $f$ is continuous at $a$, and $f(a) > 0$. Then $f(x) > 0$ for all $x$ in some interval containing $a$; more precisely, there is a number $\delta>0$ such that $f(x) > 0$ for all $x$ satisfying $|x - a| < \delta$. Similarly, if $f(a) < 0$, then there is a number $\delta>0$ such that $f(x) < 0$ for all $x$ satisfying $|x - a| < \delta$.
Proof:
Consider the case $f(a) > 0$. Since $f$ is continuous at $a$, for every $\epsilon>0$ there is a $\delta>0$ such that, for all $x$,
$\text{if} \; |x - a| < \delta, \; \text{then} \; |f(x) - f(a)| < \epsilon$ i.e., $-\epsilon < f(x) - f(a) < \epsilon$.
In particular, this must hold for $\epsilon = \frac{1}{2}f(a)$, since $\frac{1}{2}f(a) > 0$. Thus there is $\delta>0$ so that for all $x$,
$\text{if} \; |x - a| < \delta, \; \text{then} \; -\frac{1}{2}f(a) < f(x) - f(a) < \frac{1}{2}f(a)$,
and this implies that $f(x) > \frac{1}{2}f(a) > 0$. A similar proof can be given in the case $f(a) > 0$; take $\epsilon = -\frac{1}{2}f(a)$. Or one can apply the first case to the function $-f$. $Q.E.D.$
