Let $f$ and $g$ be real or complex functions having the limits$$\lim_{x\to x_0}f(x) = F \quad \mbox{and} \quad \lim_{x\to x_0}g(x) = G.$$ Then also the limit $\displaystyle\lim_{x\to x_0}f(x)g(x)$ exists and equals $FG$ .
Proof. Let $\varepsilon$ be any positivenumber. The assumptions imply the existence of the positive numbers $\delta_1,\,\delta_2,\,\delta_3$ such that
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(1)
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(2)
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(3)
According to the condition (3) we see that $$|g(x)| = |g(x)\!-\!G\!+\!G| \leqq |g(x)\!-\!G|+|G| < 1\!+\!|G|\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_3.$$ Supposing then that $0 < |x-x_0| < \min\{\delta_1,\,\delta_2,\,\delta_3\}$ and using (1) and (2) we obtain
This settles the proof.
"proof of limit rule of product" is owned by pahio.
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