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If $\lim_{x \to a}f(x)$ and $\lim_{x \to a} (f(x)g(x))$ exist, must $\lim_{x \to b} g(x)$ exist?

Let $f,g$ be two functions defined in some open interval $I$ containing a point $a$. Let us suppose that $a=b$.
Let $\epsilon>0$. We know that $\lim_{x \to a}f(x)$ and $\lim_{x \to a}(f(x)g(x))$ exist.
Therefore, let $\lim_{x \to a}f(x)=l$ and $\lim_{x \to a}(f(x)g(x))=l'$.
Since $\lim_{x \to a}f(x)=l$ and $\lim_{x \to a}(f(x)g(x))=l'$, there exists $\eta_1>0$ and $\eta_2>0$ such that:

$\forall x \in I $ \begin{array}{lcl} \mid x-a \mid \leq \eta_1 \Rightarrow \mid f(x)-l \mid \leq \\ \mid x-a \mid \leq \eta_2 \Rightarrow \mid f(x)g(x)-l' \mid \leq \epsilon\\ \end{array}

I didn't find the first line has to be smaller that and how to I reveal the $g(x)$?

Thank you in advance.

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What could go wrong if $\lim\limits_{x\rightarrow a}f(x)=0$? – David Mitra Oct 4 '12 at 17:30

$\lim\limits_{x \to b} g(x)$ not necessarily exist. If $\lim\limits_{x \to a} f(x)=0$, for example, $f(x)=\sin{x}$, $a=0$, $g(x)=\dfrac{1}{\sqrt{x}}$.

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So I can simply state that as a counter example ? – user43418 Oct 4 '12 at 17:44
In your counter example do I consider as well a=b? – user43418 Oct 4 '12 at 17:49
In case $\lim\limits_{x \to a} f(x)=0$ — yes. – M. Strochyk Oct 4 '12 at 17:50
Ok. I understood. Thank you – user43418 Oct 4 '12 at 17:51
Generally, limit behaviour of $g$ at point $b \ne a$ not must depend from its values at $a$. – M. Strochyk Oct 4 '12 at 17:58

You have

$$ \lim_{x\to a}g(x)=\lim_{x\to a}\frac{f(x)g(x)}{f(x)} $$

and applying the theorem on the limit of the ratio, if the two limits are finite and the limit of $f$ is not $0$, this becomes

$$ \frac{\lim_{x\to a}f(x)g(x)}{\lim_{x\to a}f(x)} $$

so in these hypotheses it exists.
This does not exclude that the limit could exist also if the hypotheses are not satisfied.

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Yes but that is useful for later in the proof. What element which is strictly positive should I chose such that I will have: $\mid f(x)-l \mid \leq$ "that element" – user43418 Oct 4 '12 at 17:41
@user43418: I do not want to do a proof if I can use an already known theorem. – enzotib Oct 4 '12 at 17:44

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