Theorem: If $\lim_{x\to a}\ f(x)=L$ and $\lim_{x\to a}\ g(x)=M$ then $\lim_{x\to a}(f(x)g(x))=LM$.
Proof:

Suppose $\lim_{x\to a}\ f(x)=L$ and $\lim_{x\to a}\ g(x)=M$, and consider any $\epsilon >0$. We must find a $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)g(x)-LM|<\epsilon$.

First, note that $|f(x)g(x)-LM|=|f(x)g(x)-Lg(x)+Lg(x)-LM|$

$=|g(x)(f(x)-L)+L(g(x)-M)|$

$\le|g(x)(f(x)-L)|+|L(g(x)-M)|$ by the triangle inequality

$=|g(x)||(f(x)-L)|+|L||(g(x)-M)|$.

We aim to find $\delta$ that makes each of these two terms $\le \frac{\epsilon}{2}$.

Since $\lim_{x\to a}\ g(x)=M$ we can find a $\delta_1>0$ such that if $0<|x-a|<\delta_1$ then $|g(x)-M|<\frac{\epsilon}{2(1+|L|)}$.

Also, there exists a $\delta_2>0$ such that if $0<|x-a|<\delta_2$ then $|g(x)-M|<1$, and hence $|g(x)|=|g(x)-M+M|\le |g(x)-M|+|M|<1+|M|$.

Since $\lim_{x\to a}\ f(x)=L$ we can find a $\delta_3>0$ such that if $0<|x-a|<\delta_3$ then $|f(x)-L|<\frac{\epsilon}{2(1+|M|)}$.

Take $\delta=\text{min}\{\delta_1,\delta_2,\delta_3\}$ and assume $0<|x-a|<\delta$.

Then $0<|x-a|<\delta_1,\delta_2,\delta_3$, so $|f(x)g(x)-LM|\le |g(x)||f(x)-L|+|L||g(x)-M|<(1+|M|)(\frac{\epsilon}{2(1+|M|)})+ |L|(\frac{\epsilon}{2(1+|L|)})<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.

This proves that $\lim_{x\to a}\ (f(x)g(x))=LM$.