# Help with Epsilon Delta Proof of Limit Property

My teacher was proving that if $$\lim_{x\to c} f(x)=L$$ and $$\lim_{x\to c} g(x)=M$$, then $$\lim_{x\to c} [f(x)g(x)]=LM$$.

I was super lost during the proof but I will try to recreate parts of it here from my notes.

WTS for every $$\epsilon >0$$, there's some $$\delta >0$$ s.t. $$|f(x)g(x)-LM|<\epsilon$$ whenever $$0<|x-c|<\delta$$ This part I understand.

She then says that from $$|f(x)g(x)-LM|$$ we get $$|f(x)||g(x)-M|+|M||f(x)-L|$$ This part also made sense, now this is where it gets confusing.

She stated that we have to bound the previous result and that $$|f(x)|<\max\{|L+1|,|L-1|\}$$ I am not sure what this is saying and why it is useful to us.

Then that helps us to find the $$\delta$$ such that $$|f(x)||g(x)-M|+|M||f(x)-L|<(|L|+1)|g(x)-M|+(|M|+1)|f(x)-L|$$ I am not sure how this is helping us.

Now, here I am writing exactly what she said word for word because I am very lost on this part,

Since $$\lim_{x\to c} f(x)=L$$ & $$\lim_{x\to c} g(x)=M$$,

for $$a=\frac{\epsilon}{2\max\{|L|+1,|M|+1\}}$$ (where does this even come from??) there's some $$\delta_2$$, s.t. we have $$0<|x-c|<\delta_2$$ we have $$|f(x)-L|

Take $$\delta=\min\{\delta_1, \delta_2\} < (|L|+1)\cdot a +(|M|+1)\cdot a =\frac{\epsilon}{2\max\{|L|+1, |M|+1\}}\cdot (|L|+1 + |M|+1)<\epsilon$$

Is there a better way to approach this problem? I genuinely do not understand how this proves our desired result. Thanks.

• $\lvert f(x)g(x)-LM\rvert=\lvert f(x)g(x)-Lg(x) +Lg(x)-LM\rvert$ which can be reduced using the triangle inequality. Nov 19, 2023 at 3:07

## 1 Answer

I might write this in a slightly different way, but here is how I would explain the proof.

First of all, know that a lot of the ugly-looking details come at the end once you know what specific bounds you will need. So the process of figuring out the proof is quite different from reading the complete proof.

The task at hand is to show that $$|f(x)g(x)-LM|$$ is arbitrarily small if we choose $$x$$ to be "close enough" to $$c$$. The reason this should work is because we already know $$|f(x)-L|$$ and $$|g(x)-M|$$ can be made small. Presumably then, we somehow need to rewrite $$|f(x)g(x)-LM|$$ in terms of $$|f(x)-L|$$ and $$|g(x)-M|$$.

For example, \begin{align} |f(x)g(x)-LM|&=|f(x)g(x)-Lg(x)+Lg(x)-LM|\\ &\le |f(x)g(x)-Lg(x)|+|Lg(x)-LM|\\ &=|f(x)-L||g(x)|+|L||g(x)-M|. \end{align}

Roughly speaking, this should be enough to get the result: we know that $$|f(x)-L|$$ and $$|g(x)-M|$$ can be made arbitrarily small, so we are finished if we can put some bound on $$|g(x)|$$ and $$|L|$$. Of course $$|L|$$ is already a constant, so if we ensure that $$|g(x)-M|$$ is smaller than $$\varepsilon/(2|L|+1)$$, for example, then $$|L||g(x)-M|\le |L|\varepsilon/(2|L|+1)<\varepsilon/2$$.

Note: we want each of $$|f(x)-L||g(x)|$$ and $$|L||g(x)-M|$$ to be smaller than $$\varepsilon/2$$ so that together they are less than $$\varepsilon$$. The reason we can't use $$|g(x)-M|<\varepsilon/(2|L|)$$ is that $$|L|$$ could be $$0$$.

I think your confusion comes in with the next step. We want to ensure $$|f(x)-L||g(x)|$$ is also smaller than $$\varepsilon$$, which means we need to put some bound on $$|g(x)|$$. This is slightly harder, since $$|g(x)|$$ depends on $$x$$. We can't declare that $$|f(x)-L|<\varepsilon/(2|g(x)+1)$$ and be done because $$\varepsilon/(2|g(x)+1)$$ is not a constant. Instead, the intuition is that, since $$g(x)$$ is close to $$M$$ if $$x$$ is close to $$c$$, we know that $$|g(x)|$$ can be bounded. In other words, we can guarantee $$|g(x)-M|<\varepsilon_0$$ for whatever $$\varepsilon_0>0$$ we choose. The inequality $$|g(x)-M|<\varepsilon_0$$ means that $$g(x)$$ is stuck between $$M-\varepsilon_0$$ and $$M+\varepsilon_0$$. Hence $$|g(x)|$$ is bounded by the larger of the two. I.e. $$|g(x)|<\max\{|M-\varepsilon_0|,|M+\varepsilon_0|\}$$. We just need to bound $$|g(x)|$$, so we might as well choose $$\varepsilon_0=1$$ and conclude that, for some $$\delta_0>0$$, if $$0<|x-c|<\delta_0$$, then $$|g(x)|, where $$N=\max\{|M+1|,|M-1|\}$$.

Now to wrap everything up, we want $$|f(x)-L|$$ to be smaller than $$\varepsilon/(2N)$$. Then $$|f(x)-L||g(x)|<\varepsilon/(2N)N=\varepsilon/2$$.

Finally, pick $$\delta$$ to be the smallest of all three deltas (the one we used to make $$|g(x)-M|$$ small, the one we used to bound $$|g(x)|$$, and the one we used to make $$|f(x)-L|$$ small). It follows that $$|f(x)g(x)-LM|<\varepsilon/2+\varepsilon/2=\varepsilon$$ whenever $$0<|x-c|<\delta$$.

I want to add that there is an arguably easier approach if you don't like the part about bounding $$|g(x)$$|. You can instead bound $$|g(x)|$$ using the triangle inequality again: $$|g(x)|\le |g(x)-M|+|M|$$, so $$|f(x)-L||g(x)|\le |f(x)-L||g(x)-M|+|f(x)-L||M|$$.

The proof then starts with

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

Pick $$\delta_1>0$$ so that $$|f(x)-L|<\min\{1,\varepsilon/(3|M|+1)\}$$ and $$\delta_2$$ so that $$|g(x)-M|<\min\{\varepsilon/3,\varepsilon/(3|L|+1))\}$$. If $$0<|x|<\delta$$ for $$\delta=\min\{\delta_1,\delta_2\}$$ each of the three terms above should be $$<\varepsilon/3$$.