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 $\delta_1$?
Best Answer
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 $lm$.