If $\lim\limits_{z\to z_0} f(z)=0$ and $|g(z)|
Statement: If $\lim\limits_{z\to z_0} f(z)=0$ and $|g(z)|<M$, for all $z$, with $M$ being a positive number, then we have $\lim\limits_{z\to z_0}  f(z)g(z)=0$.
I just wanted to verify my proof here:
The condition $\lim_{z \to z_0}f (z)=0$ imñlies that for all $\epsilon>0$ there exists $\delta>0$ s.t. $|f(z)-0|<\frac{\epsilon}{M+1}$ for all $|z-z_0|$.  If $\delta=\min\{\delta_{1}, \delta_{2}\}$ then for $|z-z_0|<\delta$ we have $ |f(z)||g(z)| \leq |f(z)||g(z)|<\epsilon$.  
Is this proof correct?  I think one thing I need help with is clarifying where $\delta_{2}$ comes in.  

Boundness inequality $|g(z)|<M$ needs to take place only on a neighborhood of $z_0$ (not the whole domain of definition, $D$, common to both $f$ and $g$), say for all $z\in (z_0-\delta_1, z_0-\delta_1) \cap D$, with given $\delta_1>0$.
On the other hand, given $\epsilon>0$, there is $\delta_2>0$ such that $|f(z)|<\epsilon/M$ for all  $z\in   (z_0-\delta_2, z_0+\delta_2)\cap (D\setminus \{z_0\})$.
Then: $$ |f(z)g(z)| = |f(z)||g(z)| < \epsilon/M\cdot M = \epsilon$$
for all $z\in   (z_0-\delta, z_0+\delta)\cap (D\setminus \{z_0\})$, where
$$\delta = \min (\delta_1, \delta_2).$$ 
