Limit of Product of Two Functions Given $\lim_{x\rightarrow a} f(x)=l$ and $\lim_{x\rightarrow a} g(x)=m$, I tried to prove that $lim_{x\rightarrow a} f(x)g(x)=lm$. Although the statement is simpler, proof is not obvious, and I faced some minor problems while writing the proof. Let $\epsilon >0$ be given. We have to find $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)g(x)-lm|<\epsilon$. Now, $|f(x)g(x)-lm|\leq |g(x)|.|f(x)-l|+|l|.|g(x)-m|$, and it is sufficient to make the two terms on RHS less than $\epsilon/2$ for suitable $\delta$. 
As limit of $g(x)$ is $m$, we can find $\delta_1>0$ such that 
(*) $|g(x)-m|<\frac{\epsilon}{2(1+|l|)}$ whenever $0<|x-a|<\delta_1$. 
Here we added $1$ to $l$ in the denominator since $l$ could be zero. 
For $0<|x-a|<\delta_1$, we have  $|g(x)|\leq |g(x)-m|+|m|<\frac{\epsilon}{2(1+|l|)}+|m|=C$ say. Here $C>0$. 
Now, as limit of $f(x)$ is $l$, we can find $\delta_2>0$ such that 
$|f(x)-l|<\frac{\epsilon}{2C}$ whenever $0<|x-a|<\delta_2$. 
Now one can easily show that if $\delta=min(\delta_1,\delta_2)$, then for $0<|x-a|<\delta$, both the terms $|g(x)|.|f(x)-l|$ and $|l|.|g(x)-m|$ are less than $\epsilon/2$, hence $|f(x)g(x)-lm|<\epsilon$. 
My questions are as follows:
Is it necessary to consider case $l=0$ separately? Because After choosing $\delta_1$ such that (*) holds, I tried to show that $|l|.|g(x)-m|<\epsilon/2$ for $0<|x-a|<\delta_1$, which follows easily if $l\neq 0$. Does it follow easily if $l=0$?
In the proof, should we choose $\delta_2>0$ which is less than or equal to $\delta_1$ or we can choose $\delta_2$ independent of $\delta_1$?
 A: If $|a-l|<\epsilon$ and $|b-m|<\epsilon$, then $$|ab-lm| = |a(b-m) + m(a-l)|<|a||b-m| + |m||a-l|$$
meaning that if $a$ is sufficiently close to $l$ and $b$ is close to $m$, then $ab$ will be close to $lm$.
A: Note that the following identity can simplify the proof:
$$f(x)g(x)-lm=(f(x)-l)(g(x)-m)+m(f(x)-l)+l(g(x)-m)$$
If you take $0<\varepsilon <1$ and $\delta>0$ as indicated by Michael Shi, you get immediately
$$|f(x)g(x)-lm|\leq \varepsilon(\varepsilon+|m|+|l|)\leq \varepsilon(1+|m|+|l|)$$ 
and you are done without particular cases.
A: Regarding your second question, we are using the definition of $\lim_{x \rightarrow a} f(x) = l$ and $\lim_{x \rightarrow a} g(x) = m$ to find our $\delta_1$ and $\delta_2$. It's not so much about choosing our $\delta_2$ in relation to $\delta_1$ but rather using the information we are given. In choosing our $\delta = \min (\delta_1, \delta_2)$, we are necessarily forcing both conditions $$|x - a| < \delta \leq \delta_1 \implies |f(x) - l| < \epsilon$$ and $$|x - a| < \delta \leq \delta_2 \implies |g(x) - m| < \epsilon$$ to be true. I may be misinterpreting your question though.
A: You can prove this by the following way which I feel is simpler : 
$\lim_{x\to a}f(x)=l$, so $\forall \epsilon>0,\exists \delta_1>0$ such that $0<|x-a|<\delta_1 \implies|f(x)-l|<\epsilon$. Similarly 
$\lim_{x\to a}g(x)=m$, so $\forall \epsilon>0,\exists \delta_2>0$ such that $0<|x-a|<\delta_2 \implies|g(x)-m|<\epsilon$. So, 
letting $\delta=min(\delta_1,\delta_2)$ and 0<|x-a|< $\delta$, 
$|f(x)g(x)-lm|=|f(x)g(x)-mf(x)+mf(x)-lm|=|f(x)(g(x)-m)+m(f(x)-l)|$      $$  
   \le|f(x)||g(x)-m|+m|f(x)-l|$$            $$
      \lt\epsilon(|f(x)|+m)$$  
which proves the result. $\square$  
Sorry I could not format it properly.
