Proof of product rule for limits

Question

Let

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

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

Where $L$ and $M$ are finite reals.

Then I want to prove that

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

Let $\epsilon > 0$. We need a $\delta > 0$ such that for all $x$ we have $0 < |x-a| < \delta$ implying $|f(x)g(x) - LM| < \epsilon$.

Rearrange:

$$\begin{align}|f(x)g(x)-LM|&=|f(x)g(x)-Lg(x)+Lg(x)-LM|\\ &=|g(x)(f(x)-L)+L(g(x)-M)|\\ &\le|g(x)||f(x)-L|+|L||g(x)-M| \\ &\lt|g(x)||f(x)-L|+(|L| + 1)|g(x)-M| < \epsilon\end{align}$$

Since the limits for $g(x)$ and $M$ approach the same value $M$, there exists a $\delta_1 > 0$ such that for all $x$, $0 < |x-a| < \delta_1$ implies $|g(x) - M| < \frac{\epsilon}{2(|L|+1)}$. Then:

$$\begin{align}|f(x)g(x)-LM| &\lt |g(x)||f(x)-L|+(|L|+1)|g(x)-M|\\ &<|g(x)||f(x)-L|+(|L|+1)\frac{\epsilon}{2(|L|+1)} \\ &=|g(x)||f(x)-L|+\frac{\epsilon}{2} = \epsilon \\ \end{align}$$

Since the limits for $f(x)$ and $L$ approach the same value $L$, there exists a $\delta_2 > 0$ such that for all $x$, $0 < |x-a| < \delta_2$ implies $|f(x) - L| < \frac{\epsilon}{2(|M|+1)}$. Then:

$$\begin{align}|f(x)g(x)-LM| &\lt |g(x)||f(x)-L|+\frac{\epsilon}{2} \\ &<|g(x)|\frac{\epsilon}{2(|M|+1)}+\frac{\epsilon}{2} = \epsilon \\ \end{align}$$

Now we prove $|g(x)| \leq |M|+1$:

$$|g(x)| = |g(x) - M + M| \leq |g(x) - M| + |M| \leq |M|+1$$

Subtracting $|M|$ from both sides, we see that:

$$|g(x) - M| \leq 1$$

Since the limits for $g(x)$ and $M$ approach the same value $M$, there exists a $\delta_3$ such that for all $x$, $0 < |x-a| < \delta_3$ implies $|g(x) - M| < 1$.

$$\begin{align}|f(x)g(x)-LM| &\lt |g(x)|\frac{\epsilon}{2(|M|+1)}+\frac{\epsilon}{2} \\ &< (|M|+1)\frac{\epsilon}{2(|M|+1)}+\frac{\epsilon}{2} \\ &= \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon \\ \end{align}$$

This is true granted that we set $\delta = \min(\delta_1, \delta_2, \delta_3)$.

Is my proof correct and accurate? Have I actually proved the rule?

Answer 1

Rewriting your proof:

Let $|f(x) - L| < \epsilon_1$ for $|x-a| < \delta_1$

Let $|g(x) - M| < \epsilon_2$ for $|x-a| < \delta_2$

\begin{align}|f(x)g(x)-LM|&=|f(x)g(x)-Lg(x)+Lg(x)-LM|\\ &=|g(x)(f(x)-L)+L(g(x)-M)|\\ &\le|g(x)||f(x)-L|+|L||g(x)-M| \\ &\lt|M|\epsilon_1+|L|\epsilon_2 \end{align}

Since $\epsilon_1$ and $\epsilon_2$ are arbitrarily small, the proof can end here I suppose.

Answer 2

Your proof is almost identical to the proof my calculus professor showed us in class, but I think there's one little thing you can fix to make it more rigorous. In your proof you wrote, $$|g(x)| = |g(x) - M + M| \leq |g(x) - M| + |M| \leq |M|+1$$

The last inequality isn't quite trivial, so it would be better if you write something along the lines of

$$\text{Let}\; \frac{\epsilon}{2(|L|+1)} = 1,\;\text{then}\; \exists{\delta' \gt 0} \;\text{such that}\; |x-a|\lt \delta'\; \text{implies}\; |g(x)-M|\lt 1$$

Answer 3

Let $l_{ f } = \displaystyle \lim_{ x \to a } f \left( x \right)$ and $l_{ g } = \displaystyle \lim_{ x \to a } g \left( x \right)$.

For any $\epsilon > 0$, there must be a $\delta > 0$ such that $\left| x - a \right| < \delta$ implies \begin{equation*} \left| f \left( x \right) - l_{ f } \right| < \min \left( \dfrac{ \epsilon }{ 2 \left( \left| l_{ g } \right| + 1 \right) }, 1 \right) \quad \text{and} \quad \left| g \left( x \right) - l_{ g } \right| < \dfrac{ \epsilon }{ 2 \left( \left| l_{ f } \right| + 1 \right) }. \end{equation*}

We have \begin{alignat*}{1} \left| f \left( x \right)g \left( x \right) - l_{ f } l_{ g } \right| &= \left| \left( f \left( x \right) - l_{ f } \right) l_{ g } + l_{ f } \left( g \left( x \right) - l_{ g } \right) + \left( f \left( x \right) - l_{ f } \right) \left( g \left( x \right) - l_{ g } \right) \right| \\ &\le \left| f \left( x \right) - l_{ f } \right| \left| l_{ g } \right| + \left| l_{ f } \right| \left| g \left( x \right) - l_{ g } \right| + \left| f \left( x \right) - l_{ f } \right| \left| g \left( x \right) - l_{ g } \right| \\ &< \dfrac{ \epsilon }{ 2 \left( \left| l_{ g } \right| + 1 \right) } \left| l_{ g } \right| + \left| l_{ f } \right| \dfrac{ \epsilon }{ 2 \left( \left| l_{ f } \right| + 1 \right) } + \left| f \left( x \right) - l_{ f } \right| \left| g \left( x \right) - l_{ g } \right| \\ &< \dfrac{ \epsilon }{ 2 \left( \left| l_{ g } \right| + 1 \right) } \left| l_{ g } \right| + \left| l_{ f } \right| \dfrac{ \epsilon }{ 2 \left( \left| l_{ f } \right| + 1 \right) } + \left| g \left( x \right) - l_{ g } \right| \\ &< \dfrac{ \epsilon }{ 2 \left( \left| l_{ g } \right| + 1 \right) } \left| l_{ g } \right| + \left| l_{ f } \right| \dfrac{ \epsilon }{ 2 \left( \left| l_{ f } \right| + 1 \right) } + \dfrac{ \epsilon }{ 2 \left( \left| l_{ f } \right| + 1 \right) } \\ &= \dfrac{ \epsilon }{ 2 \left( \left| l_{ g } \right| + 1 \right) } \left| l_{ g } \right| + \left( \left| l_{ f } \right| + 1 \right) \dfrac{ \epsilon }{ 2 \left( \left| l_{ f } \right| + 1 \right) } \\ &= \dfrac{ \epsilon }{ 2 \left( \left| l_{ g } \right| + 1 \right) } \left| l_{ g } \right| + \dfrac{ \epsilon }{ 2 } \\ &= \dfrac{ \epsilon }{ 2 } \cdot\dfrac{ \left| l_{ g } \right| }{ \left| l_{ g } \right| + 1 } + \dfrac{ \epsilon }{ 2 } \\ &< \dfrac{ \epsilon }{ 2 } + \dfrac{ \epsilon }{ 2 } \\ &= \epsilon. \end{alignat*}

Hence $\left|x - a\right| < \delta$ implies $\left| f \left( x \right) g \left( x \right) - l_{ f } l_{ g } \right| < \epsilon$ and thus $\displaystyle \lim_{ x \to a } f \left( x \right) g \left( x \right) = l_{ f } l_{ g }$.

This is the proof from "Calculus" by Michael Spivak with some extra steps added when deriving the inequality.