Limit of a product of two functions is $0$ 
If $f(x)$ and $g(x)$ two real valued functions such that $\lim\limits_{x\to c}f(x)$ exists and is finite, $g(x)$ is bounded and $$\lim_{x\to c}f(x) g(x) =0$$
then prove that $\lim\limits_{x\to c}f(x)=0$ or $\lim\limits_{x\to c}g(x)=0$ or both.

My try: let $\lim\limits_{x\to c}f(x)\neq 0$ then
$$\lim\limits_{x\to c} g(x) =\frac{\lim\limits_{x\to c}f(x) g(x)   }{\lim\limits_{x\to c}f(x)   }$$ and hence we have $$\lim_{x\to c} g(x) =0  $$
What if $\lim\limits_{x\to c}g(x)\neq 0$?
Thank you.
 A: Let  $\displaystyle \lim_{x \to c}f(x)=l\neq\,0$. Write
$f(x)g(x)=(f(x)-l+l)g(x)=(f(x)-l)g(x)+lg(x)$. It is clear that :
$|(f(x)-l)g(x)|\leq\,M\,|f(x)-l|$ and clearly $\displaystyle \lim_{x \to c}(f(x)-l)g(x)=0$.
Therefore $\displaystyle \lim_{x \to c}l\,g(x)=0$. Thus $\displaystyle \lim_{x \to c}\,g(x)=0$.
Now if $l=0$ then $\displaystyle \lim_{x \to c}f(x)=0$.
And clearly we can have both $\displaystyle \lim_{x \to c}\,f(x)=0$ and
$\displaystyle \lim_{x \to c}\,g(x)=0$.
A: If $\lim_{x\to c} f(x) = 0$ then we are done.
So suppose $\neg(\lim_{x\to c} f(x) = 0).\ $ Since $\lim_{x\to c} f(x)\ $ exists, let $b = \lim_{x\to c} f(x)\neq 0.$
By definition, $ \exists\ u>0$ such that $\ f(w) \in \left(\frac{b}{2}, \frac{3b}{2}\right)\quad \forall\ w\in (c-u,\ c+u).$
Suppose further that $\neg\left(\lim\limits_{x\to c}g(x)=0\right)\ $ and let $\ \varepsilon > 0.\ $
Then there exists $d>0\ $ such that for every $\ \delta>0,\ \exists\ \delta'\in (c-\delta,\ c+\delta)\ $ such that $\ \vert g(\delta') \vert > d. $
In particular, for every $\ \delta>0,\ \exists\ \delta'\in (c-u,\ c+u)\ $ such that $\ \vert g(\delta') \vert > d. $
Thus, for every $\ \delta>0,\ \exists\ \delta'\in (c-u,\ c+u)\ $ such that $\ \vert f(\delta')g(\delta') \vert > \frac{db}{2},\ $ proving that $\lim\limits_{x\to c}f(x)g(x)\neq0.$
A: There are two cases, depending on $\lim_{x\to c} f(x)$ (which exists and is finite):

*

*If $\lim_{x\to c} f(x)=0$, then $\lim_{x\to c} f(x)=0$.


*If $\lim_{x\to c} f(x)\ne0$, then in your try, you already proved that $\lim_{x\to c} g(x) = 0$:
$$
\lim_{x\to c} g(x) = \frac{\lim_{x\to c} f(x)g(x)}{\lim_{x\to c} f(x)}
= 0
$$
Either way, $\lim_{x\to c} f(x)=0$ or $\lim_{x\to c} g(x)=0$ (or both).
If you really want to know what if $\lim_{x\to c}g(x)\ne 0$ or this limit doesn't exist, by the contrapositive of your try, then $\lim_{x\to c}f(x)=0$.
A: There is a theorem: if $\displaystyle\lim_{x\to c} b(x)=b$ and $\displaystyle\lim_{x\to c} a(x)=a\neq 0,$ then $$\lim_{x\to c}{b(x)\over a(x)}={b\over a}$$
Assume $\displaystyle\lim_{x\to c} f(x)=a\neq 0.$ Then substituting $a(x)=f(x)$ and $b(x)=f(x)g(x)$ we get $b=0$ and $$\lim_{x\to c}g(x)=\lim_{x\to c}{b(x)\over a(x)}={b\over a}=0$$
