Stuck on trying to prove product rule for limits

Question

Let

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

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

Then I want to prove that

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

Let $\epsilon > 0$. Then I want to prove that there exists a $\delta > 0$ such that for all $x$ we have $0 < |x-a| < \delta$ implying $|f(x)g(x) - LM| < \epsilon$.

I did this:

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

There exists a $\delta_1 > 0$ such that for all $x$, $0 < |x-a| < \delta_1$ implies $|g(x) - M| < \frac{\epsilon}{2|L|}$. Then:

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

I figure for symmetry's sake I should be trying to do the same thing for the other side.

There exists a $\delta_2 > 0$ such that for all $x$, $0 < |x-a| < \delta_2$ implies $|f(x) - L| < \frac{\epsilon}{2|M|}$. Then:

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

Now if I can prove that $|g(x)| \leq |M|$ I could cancel that stuff out and the inequality would hold.

$$|g(x)| = |g(x) - M + M| \leq |g(x) - M| + |M| \leq |M| $$ But then this implies $|g(x) - M| \leq 0 $, but I can't prove it equals zero since this would require $\epsilon = 0$ which is a contradiction.

1. Where did I screw up?

2. Do I need to explicitly state that $L$ and $M$ exist at the beginning? Do I have to take into account situations where they are infinity? Or zero (in case I ever need to divide by $M$ or $L$ anywhere)?

3. The lines where I declared $\delta_1, \delta_2$ to get those absolute-value statements less than some convenient manipulation of $\epsilon$, why am I allowed to do that? It seems weird to me that I can just declare that there exists a $\delta_1$ that makes $|g(x) - M| < \frac{\epsilon}{2|L|}$. Like, do I have to prove that too? How do I know that's even valid? What stops me from similarly declaring that the $\delta > 0$ exists at the beginning to make $|f(x)g(x) - LM| < \epsilon$ true?

Answer 1

So by your definitions we have that $\forall \epsilon >0, \exists \delta_1, \delta_2 >0$ such that $$|f(x) - L| < \epsilon, \forall x\in \mathcal{D}(f) \ \text{with}\ 0<|x-a|<\delta_1,$$ and $$|g(x) - M| < \epsilon, \forall x\in \mathcal{D}(g) \ \text{with}\ 0<|x-a|<\delta_2,$$ What you continued to do is also sufficient, so following your results we have that

$$|f(x)g(x) - LM| < |g(x)||f(x)-L| + |L||g(x) - M| < |g(x)|\epsilon + |L|\epsilon$$

All that remains to do is bound this $|g(x)|$ part. Take $\epsilon =1$ in the definition of the limit for $g$, then $$|g(x) - L| <1 \implies |g(x)| < |L| +1,$$ which holds if we choose $\delta_3 = \delta_2(1)$. Thus: $$|f(x)g(x) - LM| < (|L|+1)|f(x)-L| + |L||g(x) - M| < (|L|+1)\epsilon + |L|\epsilon$$ holds if we take $\delta = \min \{\delta_1, \delta_2, \delta_3\}$. To make this look more pleasing, you could alter the bounds in the definition of limits for $f$ and $g$ by introducing $L$.

Answer 2

It is recommended to control like $|g(x)-M|<\dfrac{\epsilon}{2(|L|+1)}$ and then $|L||g(x)-M|\leq\dfrac{|L|\epsilon}{2(|L|+1)}<\dfrac{\epsilon}{2}$ because $L$ could be zero.

Now control one $\delta'>0$ such that $|g(x)-M|<1$ (or any other constant) then $|g(x)|\leq|g(x)-M|+|M|<1+|M|$, so by taking minimum $\delta=\min\{\delta',\delta_{1},\delta_{2}\}$ we have $|g(x)||f(x)-L|<(1+|M|)|f(x)-L|<\dfrac{(1+|M|)\epsilon}{2}$ for all $0<|x-a|<\delta$.

Answer 3

You have $$|f(x)g(x)-LM|\le|g(x)||f(x)-L|+|L||g(x)-M|$$

you have taken care of $$|L||g(x)-M|< \epsilon/2$$

For

$$|g(x)||f(x)-L|$$Make $$|g(x)|<M+1$$ and $$ |f(x)-L|< \frac {\epsilon}{2(M+1)}$$

by choosing your $\delta_1$ and $\delta_2$ small enough.