I was trying to understand the definition of the limit since it's literally the bedrock of Calculus. The typical idea of convergence is that for some limiting value $L$, the value of a function can always be as close to $L$ as we want. The $\epsilon-\delta$ definition states that there can only be a single limiting value (somewhat implicitly). In the case of something like a jump discontinuity, the limit doesn’t exist.
Suppose we have a function $f(x)$ that’s differentiable everywhere and that $\lim_{x\to a}f(x)=L$. Consider the typical setup for this limit but with one addition: for any $\epsilon\gt0$ let $$0\lt\xi\lt\epsilon$$ Note that since $\xi\gt0$, $L\neq L+\xi$. I get that the limit has been defined to only have a single limiting value, but the definition seems to have an ambiguity.
Doesn't the limit converge to both $L$ and $L+\xi$ since we can always choose a $\xi<\epsilon$ in the limiting process?
Best Answer
I don't agree that "The $\varepsilon$−$\delta$ definition states that there can only be a single limiting value (somewhat implicitly)." However we can prove that the limit is unique. Suppose that $L$ and $M$ are both limits of $f$ as $x\to a$. Then for each $\varepsilon$ we can choose a $\delta$ such that both $|f(x)-L|<\varepsilon/2$ and $|f(x)-M|<\varepsilon/2$. By the triangle inequality we now get $|L-M|\leq |f(x)-L|+|f(x)-M|<\varepsilon/2+\varepsilon/2=\varepsilon$. So what we have proved is that the real number $|L-M|$ is less than any $\varepsilon$. The only such real number is $0$ so $|L-M|=0$ and $L=M$.
Now, the problem with your proof that $f$ converges to both $L$ and $L+\xi$ is that you are changing the real number after we give you an $\varepsilon$. What I'm trying to say is that for $f$ to approach a real number $L$, you have to specify that $L$ and then choose a $\delta$ such that $|f(x)-L|<\varepsilon$. You cannot keep changing your $L$.
Also, for a limit of a function to exist, it does not need to be differentiable, but only continuous. If it is differentiable however, it is also continuous.