0
$\begingroup$

Suppose $\lim\limits_{x \to a} f(x) = L$ and $\lim\limits_{x \to a} g(x) = M$. I'm trying to understand an intuitive argument for the fact that $\lim\limits_{x \to a} [f(x) + g(x)] = L + M$, that doesn't use $\epsilon$-$\delta$.

The argument goes: as $x \to a$, $f(x)$ is close to $L$ and $g(x)$ is close to $M$. So write $f(x) = L + \epsilon_1$ and $g(x) = M + \epsilon_2$. As $x \to a$, $\epsilon_1, \epsilon_2 \to 0$, so $f(x) + g(x) \to L + M$.

This argument seems circular to me, using the fact that the limit of a sum of errors equals the sum of the limits, unless I can just take for granted that two arbitrarily small things added together yields another arbitrarily small thing.

$\endgroup$
2
  • 2
    $\begingroup$ If you want an intuitive reason, then is there any reason you're not satisfied by just thinking of it as what the values approach? Otherwise if you do want something more rigorous the $\epsilon - \delta$ proof for this really isn't that bad. $\endgroup$ – Stephen Donovan Jun 17 at 21:14
  • $\begingroup$ It’s an intuitive argument. Intuitively, if $\epsilon_1,\epsiolin_2$ are small, then their sum is small. You can make it rigorous, but intuitive is not rigorous. $\endgroup$ – Thomas Andrews Jun 17 at 21:35
2
$\begingroup$

The $\epsilon$-$\delta$ definition of a limit can be explained in a somewhat intuitive way: we say that $\lim_{x\to a}f(x)=L$ if you can get $f(x)$ as close to $L$ as you want by keeping $x$ close enough to $a$. (Key part: given how close you want to be to $L$, then you figure out how close you need to be to $x$. Without writing more formally, it can be easy to get this backwards. I like to think of limits as little machines that tell you how to approximate things to a certain precision.)

To get $f(x)+g(x)$ as close as $L+M$ as you might want, let's take the allowed error and divide it into two parts. For the first part of the allowed error, since $\lim_{x\to a}f(x)=L$ we can make $f(x)$ be that close to $L$ for some $x$ near enough to $a$. Similary, for the second part of the allowed error, since we can make $g(x)$ be that close to $M$ for some $x$ near enough to $a$. If we keep $x$ close enough to $a$ (by staying as close as necessary for each of these two parts we considered) then $f(x)+g(x)$ will stay within the allowed error of $L+M$. Since we showed we can meet any allowed error, we can say $\lim_{x\to a}(f(x)+g(x))=L+M$.

The argument you gave is circular, by the way. You essentially turned $\lim_{x\to a}f(x)$ into $\lim_{x\to a}(f(x)-L)$, writing $\epsilon_1$ for $f(x)-L$, and then the core of your argument is that $\lim_{x\to a}(\epsilon_1+\epsilon_2)=\lim_{x\to a}\epsilon_1+\lim_{x\to a}\epsilon_2=0+0=0$.

$\endgroup$
1
  • $\begingroup$ See @Henry's answer for this argument written more formally. $\endgroup$ – Kyle Miller Jun 17 at 21:28
3
$\begingroup$

The better $\epsilon-\delta$ argument is that for any $\epsilon>0$

  • $\lim\limits_{x \to a} f(x) = L$ means there is a $\delta_f$ where $|x-a|\lt \delta_f \implies |f(x)-L| \lt \frac\epsilon 2$
  • $\lim\limits_{x \to a} g(x) = M$ means there is a $\delta_g$ where $|x-a|\lt \delta_g \implies |g(x)-M| \lt \frac\epsilon 2$

So letting $\delta = \min(\delta_f, \delta_g)$ you have $$|x-a|\lt \delta \implies |f(x)+g(x)-(L+M)| \le |f(x)-L| +|g(x)-M| \lt \epsilon$$ meaning $\lim\limits_{x \to a} f(x) +g(x)= L+M$

This is not circular, and it is saying intuitively we can add limits because we can get close enough to each limit at the same time so that when adding together the differences, the sum can be arbitrarily small.

$\endgroup$
1
$\begingroup$

Your last point is correct. Sticking with the intuitive intent of this argument, it is always assumed that adding together two arbitrarily small values results in another arbitrarily small value. If you can make any two values "as small as you want," as the word "arbitrarily" implies, then you can make their sum as small as you desire be making $\epsilon_1,\epsilon_2$ small enough.

$\endgroup$
2
  • $\begingroup$ This is helpful, thank you. Is the same thing true for the product of an arbitrarily small value and a constant? $\endgroup$ – Stanley Smith Jun 17 at 21:23
  • 1
    $\begingroup$ Basically the same idea. $\endgroup$ – Jake Freeman Jun 17 at 21:23
1
$\begingroup$

If $\lim_{x \to a}f(x)=L$ and $\lim_{x \to a}g(x)=M$, then by definition we can make $|f(x)-L|$ as small as we like by requiring that $x$ is sufficiently close to $a$. The same is true for $|g(x)-M|$. More explicitly, for any positive $\varepsilon_1$, and for any positive $\varepsilon_2$, there is a number $\delta$ such that $$ 0<|x-a|<\delta \implies |f(x)-L|<\varepsilon_1 \text{ and } |g(x)-L|<\varepsilon_2 \, . $$

The triangle inequality tells us $$ |f(x)-L|<\varepsilon_1 \text{ and } |g(x)-M|<\varepsilon_2 \implies \left|\bigl(f(x)+g(x)\bigr)-\bigl(M+N\bigr)\right|<\varepsilon_1+\varepsilon_2 \, . $$ Crucially, this means that the distance between $f(x)+g(x)$ and $M+N$ is controlled by the values of $\varepsilon_1$ and $\varepsilon_2$.

Now, suppose I wanted to ensure that $\left|\bigl(f(x)+g(x)\bigr)-\bigl(M+N\bigr)\right|$ be smaller than some positive $\varepsilon$. Then, I can simply need to pick values of $\varepsilon_1$ and $\varepsilon_2$ such that $\varepsilon_1+\varepsilon_2\le\varepsilon$, and the result follows. For instance, we could set $\varepsilon_1=\dfrac{\varepsilon}{2}$ and $\varepsilon_2=\dfrac{\varepsilon}{2}$. Then, $$ |f(x)-L|<\frac{\varepsilon}{2} \text{ and } |g(x)-M|< \frac{\varepsilon}{2} \implies \left|\bigl(f(x)+g(x)\bigr)-\bigl(M+N\bigr)\right|<\varepsilon \, . $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.