# Is this epsilon-delta proof sufficient?

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?

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.

• How do you know that $f$ and $g$ are continuous? Is it because $f$ and $g$ differ by a finite number of points? Commented Aug 27, 2022 at 20:01
• At no moment I assumed that $f$ or $g$ is continuous. Why do you think otherwise? Commented Aug 27, 2022 at 20:34
• 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. Commented Aug 27, 2022 at 20:40
• It was a typo. I've edited my answer. Please read it again. Commented Aug 27, 2022 at 20:49
• "And if it turns out that $c\neq a$, take...". Should it not be if it turns out that $c=a$? Commented 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).

• 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 $a=c$. Commented Aug 27, 2022 at 19:59
• 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
– paf
Commented Aug 27, 2022 at 23:18
• @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) Commented 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.