Is this epsilon-delta proof sufficient?

Question

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."

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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,

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

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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?

Answer 1

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.

Answer 2

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).

Answer 3

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.