All the proofs I've seen so for the limit of a product of functions equaling the product of the limits are based on the following:

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 $FG$.

Let $\varepsilon$ be any positive number.\, The assumptions imply the existence of the positive numbers $\delta_1,\,\delta_2,\,\delta_3$ such that \begin{align} |f(x)-F| < \frac{\varepsilon}{2(1+|G|)}\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_1\tag{1} \end{align} \begin{align} |g(x)-G| < \frac{\varepsilon}{2(1+|F|)}\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_2\tag{2} \end{align} \begin{align} |g(x)-G| < 1\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_3\tag{3} \end{align} 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 \begin{align*} |f(x)g(x)-FG|\;& = |f(x)g(x)-Fg(x)+Fg(x)-FG|\\ & \leqq |f(x)g(x)\!-\!Fg(x)|+|Fg(x)\!-\!FG|\\ & = |g(x)|\cdot|f(x)\!-\!F|+|F|\cdot|g(x)\!-\!G|\\ & < (1\!+\!|G|)\frac{\varepsilon}{2(1\!+\!|G|)}+(1\!+\!|F|)\frac{\varepsilon}{2(1\!+\!|F|)}\\ & = \varepsilon \end{align*} This settles the proof.

But after having a go myself, I came up with the following:

Let $\varepsilon$ be any positive number. The assumptions imply the existence of the positive numbers $\delta_1,\,\delta_2$ such that \begin{align} |f(x)-F| < {\varepsilon}\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_1\tag{1} \end{align} \begin{align} |g(x)-G| < {\varepsilon};\;\mbox{when}\;\;0 < |x-x_0| < \delta_2\tag{2} \end{align}

Supposing then that\, $0 < |x-x_0| < \min\{\delta_1,\,\delta_2\}$\, and using (1) and (2) we obtain \begin{align*} |f(x)g(x)-FG|\;& = |(f(x) - F)(g(x) - G) + G(f(x) - F) + F(g(x) - G)|\\ & \leq |f(x) - F||g(x) - G| + |G||f(x) - F| + |F||g(x) - G|\\ & < \varepsilon^2 + \varepsilon(|L| + |G|)\\ & = \varepsilon' \end{align*}

where $\varepsilon'$ is any postive number, giving $\varepsilon = -1/2(|F|+|G|) +1/2\sqrt{(|F| + |G|)^2 + 4\varepsilon'}$

This settles the proof.

Is my proof rigorous enough?

  • 2
    $\begingroup$ You'll need to check that $\epsilon$ is positive for all $\epsilon' >0$. $\endgroup$
    – Braindead
    Jun 19, 2015 at 2:18
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    $\begingroup$ Formally speaking, no. The technical definition is that for every $\epsilon > 0$, there exists $\delta > 0$ such that if $|x - x_{0}| < \delta$, then $|f(x)g(x) - FG| < \epsilon$. For the purposes of an analysis class, you'd need to do the problem as it was written in the first part. That having been said, when you're finished with analysis, you'll probably be able to get away with writing it the second way. $\endgroup$
    – AJY
    Jun 19, 2015 at 2:49
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    $\begingroup$ While your approach is OK, it feels more convoluted compared to the standard proof. In analysis proofs it is important to think in terms of inequalities instead of formulas which connect $\delta$ and $\epsilon$. You should be able to show the existence of a $\delta > 0 $ based on arbitrary $\epsilon > 0$ without specifying the exact dependence of $\delta$ on $\epsilon$ via a formula. For example in your last line you should say that given any $\epsilon' > 0$ it is possible to find an $\epsilon > 0$ such that $\epsilon^{2} + k\epsilon < \epsilon'$ where $k$ is a positive constant. $\endgroup$
    – Paramanand Singh
    Jun 19, 2015 at 5:07
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    $\begingroup$ If $\epsilon < 1$ then $\epsilon^{2} < \epsilon$ and hence $\epsilon^{2} + k\epsilon < (1 + k)\epsilon$ and this would be less than $\epsilon'$ if $\epsilon < 1$ and $\epsilon < \epsilon'/(1 + k)$. Thus given any $\epsilon' > 0$ we can choose an $\epsilon > 0$ such that $\epsilon < 1, \epsilon < \epsilon'/(1 + k)$ and this based on this $\epsilon$ choose your $\delta_{1}, \delta_{2}$ and $\delta$ can be taken as minimum of $\delta_{1}, \delta_{2}$. And yes you should interchange the role of symbols $\epsilon$ and $\epsilon'$ to make it like a textbook formal proof. $\endgroup$
    – Paramanand Singh
    Jun 19, 2015 at 5:11
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    $\begingroup$ As long as you are thinking of formulas connecting various epsilons with deltas you are working in "Algebra" mode. You need to come out of that mode and think in terms of inequalities. Algebra will help in simplification of certain expressions and should act as only a tool when you work in "Analysis" mode. $\endgroup$
    – Paramanand Singh
    Jun 19, 2015 at 5:12

2 Answers 2


Yes, the second proof is rigorous enough. But for clarity, it should add at the end:

Hence for any positive $\varepsilon'$, there exists a positive $\varepsilon < \varepsilon'$ that satisfies conditions (1) and (2).

This is important because you need $\varepsilon' \rightarrow 0 \Rightarrow\varepsilon \rightarrow 0$


You can simplify it if you assume that $\varepsilon < 1$.

Then, from $|f(x)g(x)-FG| < \varepsilon^2 + \varepsilon(|F| + |G|) $ (you wrote $|L|$ instead of $|F|$) you get $|f(x)g(x)-FG| < \varepsilon(1+|F| + |G|) $ so you can just choose $\varepsilon \lt \dfrac{\min(1, \varepsilon')}{1+|F| + |G|} $.


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