Question: **Theorem:**

$$\lim\_{x \to a} f(x) = L$$

$$\lim\_{x \to a} g(x) = M$$

Then:

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

Obviously,

$$|f(x) - L| < \epsilon$$

$$|g(x) - M| < \epsilon$$ 

But multiplying these together doesnt get the desired:

$$|f(x)g(x) - LM| < \epsilon$$

**Please, HINTS only!**

Answer: 
Hint:

$$\begin{align}|f(x)g(x)-LM|&=|f(x)g(x)-Lg(x)+Lg(x)-LM|\\&=|g(x)(f(x)-L)+L(g(x)-M)|\\&\le|g(x)||f(x)-L|+|L||g(x)-M|\end{align}$$
Prove it for the special case where the two functions are infinitesimal as $x\to a$ that is $\lim\_{x\to a}f(x)=0$

Prove the theorem for such special functions then deduce the theorem.