I am currently reading "Calculus a Rigorous First Course" by Daniel J. Velleman. I am on the exercise set of section 2.4 problem 29, and it states the following.

"Suppose that $f$ and $g$ are functions that agree on all values except one. In other words, there is some number $c$ such that for all $x \neq c$, $f(x) = g(x)$, but $f(c) \neq g(c) $. Show that for every number a, $\lim_{x \rightarrow a}f(x)=\lim_{x \rightarrow a}g(x)$, where we interpret this equation to mean that either both limtis are defined and they are equal, or both limits are undefined."

My attempt:

(Let $\lim_{x \rightarrow a}f(x)=\lim_{x \rightarrow a}g(x)=L$)

Proof. Supppose $\epsilon > 0$. Let $\delta=\min(\delta_1, \delta_2)$, such that if $0<|x-a|<\delta_1$ then $|f(x) - L| < \epsilon/2$. Similarly, if $0<|x-a|<\delta_2$ then $|g(x)-L|<\epsilon/2$. Now, suppose $0<|x-a|<\delta$. Because $\delta \leq \delta_1$ then $0<|x-a|<\delta_1$ which means that $|f(x)-L|< \epsilon /2$. Similarly, because $\delta \leq \delta_2$ then $0<|x-a|<\delta_2$, so $|g(x)-L|< \epsilon/2$. Therefore,

$$|f(x) - L+g(x)-L)| \leq |f(x)-L| + |g(x)-L| < \epsilon/2 + \epsilon/2 = \epsilon. $$

My doubt

I do not see how my proof has anything to do with the condition $f(c) \neq g(c)$. Also, I would like to add an additional note left on the problem, which states

"(Note: We could state this result more informally by saying the value of the limit $\lim_{x \rightarrow a}f(x)$ will not be affected if we change the value of $f(c)$, for any number $c$. This may seem paradoxical: it suggests that none of the values of the function $f$ are relevant to the value of $\lim_{x \rightarrow a}f(x)!)$"

How is the factorial of $f(x)$ relevant? Or this some sort of typo?


3 Answers 3


You proved that$$\lim_{x\to a}f(x)=L\wedge\lim_{x\to a}g(x)=L\implies\lim_{x\to a}f(x)+g(x)=2L.$$But that is not what you were supposed to have proved.

Suppose that $\lim_{x\to a}f(x)=L$. If it turns out that $a=c$, then this is exactly the same thing as proving that $\lim_{x\to a}g(x)=L$, since the value that $f$ and $g$ take at $c$ doesn't matter and $f(x)= g(x)$ when $x\ne c \wedge c=a$.

And if it turns out that $c\ne a$, take $\varepsilon>0$ and take $\delta>0$ such that$$0<|x-a|<\delta\implies|f(x)-L|<\varepsilon.$$Now, let $\delta'=\min\{\delta,|c-a|\}$. Then $0 <|x-a|<\delta'$ implies two things:

  • $|x-a|<\delta$;
  • $x\ne c$.

And therefore $|g(x)-L|=|f(x)-L|<\varepsilon$.

Now, if the limit $\lim_{x\to a}f(x)$ doesn't exist, then the limit $\lim_{x\to a}g(x)$ cannot exist, since otherwise, by the previous argument, if it did, then them limit of $f$ at $a$ would also exist.

  • $\begingroup$ How do you know that $f$ and $g$ are continuous? Is it because $f$ and $g$ differ by a finite number of points? $\endgroup$ Aug 27, 2022 at 20:01
  • $\begingroup$ At no moment I assumed that $f$ or $g$ is continuous. Why do you think otherwise? $\endgroup$ Aug 27, 2022 at 20:34
  • $\begingroup$ because you’ve stated that $|f(x) - f(a)|< \epsilon$, which is only true if $\lim_{x \rightarrow a} f(x) = f(a)$ but I don’t see how this is necessarily the case. $\endgroup$ Aug 27, 2022 at 20:40
  • $\begingroup$ It was a typo. I've edited my answer. Please read it again. $\endgroup$ Aug 27, 2022 at 20:49
  • $\begingroup$ "And if it turns out that $c\neq a$, take...". Should it not be if it turns out that $c=a$? $\endgroup$ Aug 27, 2022 at 20:55

Your proof can't be correct: you begin by assuming that the two limits are equal but this is what you want to prove! You should begin by assuming that $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} g(x) = L'$, without knowing if $L=L'$ or not. Then, write the definitions with epsilon and delta and use the fact that for all $x\neq c$, $f(x)=g(x)$ (you didn't use it in your attempt).

Finally, the exclamation mark is simply an exclamation mark (this should be surprising), not a factorial sign (which is only valid for nonnegative integers).

  • $\begingroup$ Because $f$ and $g$ differ only by a finite amount of points, does this mean that both $f$ and $g$ are continuous? If so, I would only have to argue for the case when $a=c$. $\endgroup$ Aug 27, 2022 at 19:59
  • $\begingroup$ I can't see any information about continuity of $f$ or $g$ in the information you provided up to now thus we can't know if they are continuous or not $\endgroup$
    – paf
    Aug 27, 2022 at 23:18
  • 1
    $\begingroup$ @BryanBusby - consider the case where $f(x)=1$ when $x$ is rational and $f(x)=0$ when $x$ is irrational. Then consider $g(x)$ defined similarly except that $g(0)=0$. Neither function is continuous anywhere, but they are equal to each other everywhere except $x=0$. (Note that both functions can be described meaningfully - $f(x)$ is the rational indicator function. $g(x)$ is a function that indicates if the inverse of $x$ is rational) $\endgroup$
    – Glen O
    Aug 28, 2022 at 3:20

About the comment ending "... This may seem paradoxical: it suggests that none of the values of the function f are relevant to the value of $lim_{x→a}f(x)$", it has to do with the structure of the real numbers. Between any two real numbers, there are infinite number of other real numbers. So for any $a\neq c$, we can simply further restrict $\delta < | c-a |$, thereby excluding c from the $\epsilon$ neighborhood of a.


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