Limit Product Law : Suppose $\lim_{x\to a}f(x) = L$ and $\lim_{x\to a}g(x)=M$,Then

$$\lim_{x \to a}f(x)g(x) = LM$$

Proof: Given any $\xi$ we need to find a $\delta$ such that



Here we use an algebratic trick,

$$|f(x)g(x)- LM| < \xi = |f(x)g(x) -f(x)M+f(x)M - LM| \\ = |f(x)(g(x)-M)+M(f(x)-L)|\\ \leq |f(x)(g(x)-M)|+|M(f(x)-L)|\\ = |f(x)||(g(x)-M)|+|M||(f(x)-L)|$$

Since $\lim_{x\to a}f(x)=L$, there is a value $\delta_{1}$, so that $0<|x-a|<\delta_{1}$ implies $|f(x) - L| < |\xi /(2M)|$ (此$\xi$是$|f(x)g(x)- LM| < \xi$中的$\xi$) impies $|M||(f(x)-L)| < \xi/2$

If we can make $|f(x)||(g(x)-M)| < \xi /2$， then we’ll be done.

Here we need another trick. We can find a $\delta_{2}$ so that $|x-a|< \delta_{2}$ implies that $|f(x)-L|<1$,meaning that $L -1 < f(x) < L+1$.This means that $|f(x)| < N$, wher N is either $|L - 1|$ or $|L+1|$, depending on whether L is negative or positive. The important point is N doesn’t depend on X. Finally, we know that ther is a $\delta_{3}$ so that $0<|x-1|<\delta_{3}$ implies $|g(x)-M| < \xi/(2N)$.Now we’re ready to put everything together.Let $\delta$ be the smallest of $\delta_{1}, \delta_{2}, \delta_{3}$, Then $|x-a| < \delta$ implies that（意思就是，当$\delta$取$\min\delta_{1}, \delta_{2}, \delta_{3}$时，以上三个式子都成立，即:

${\rm I}.\ 0<|x-a|<\delta_{1}\ implies\ |f(x) - L| < |\xi /(2M)|$
${\rm II}.\ 0<|x-a|< \delta_{2}\ implies\ |f(x)- L|<1$
${\rm III}.\ 0<|x-a|<\delta_{3}\ implies\ |g(x)-M| < \xi/(2N)$

$$|f(x)g(x)- LM| < |f(x)||(g(x)-M)|+|M||(f(x)-L)| = \xi /2+ \xi/2 = \xi$$