Arriving at the value of Limits

Question

According to the formal definition of limits,

Let $f(x)$ be a function defined on an open interval $D$ that contains $c$, except possibly at $x=c$. Let $L$ be a number. Then we say that

$$\lim_{x \to c} f(x) = L$$ if, for every $\varepsilon>0$, there exists a $\delta>0$ such that for all $x\in D$ with

$$0 < \left| x - c \right| < \delta$$ we have

$$\left| f(x) - L \right| < \varepsilon$$

Suppose now we have a function $f(x)=x$. Now if we want to calculate the limit at 0, we just replace x by the value 0. What decides that it is correct and is the only possible limit? We could as easily have chosen 1 as our limit still satisfy the formal definition by staying within the error limits. Why is this so?

If you open any calculus book you'll find the statement that if $$\mathop{\forall}{\varepsilon >0}\mathop{\exists}{\delta >0}\mathop{\forall}{x\in D}\left(0<|x-c|<\delta\implies |f(x)-c|<\varepsilon\right)$$ and $$\mathop{\forall}{\varepsilon >0}\mathop{\exists}{\delta >0}\mathop{\forall}{x\in D}\left(0<|x-c'|<\delta\implies |f(x)-c'|<\varepsilon\right),$$ then $c=c'$. — Git Gud, Dec 23 '14 at 01:09
Note that the ability to replace the limit at $x$ with the value $f(x)$ means that the function is continuous at $x$. — Mark Bennet, Dec 24 '14 at 07:28
Does this answer your question? Does there exist an epsilon-delta proof of the limit for all polynomials? — , Oct 08 '23 at 21:34

score 1 · Answer 1 · answered Dec 23 '14 at 01:56

We could as easily have chosen 1 as our limit still satisfy the formal definition by staying within the error limits.

Um, no. Set $\epsilon = \frac{1}{2}$. Let $\delta > 0$ be given, and set $x=\min\left\{\frac{\delta}{2},\frac{1}{2}\right\}$. Then $|x| < \delta$ but $$|f(x) - 1| = |x-1| \geq \frac{1}{2}$$

We have shown there exists some $\epsilon >0$ such that for all $\delta > 0$ there exists $x$ with $|x-0| < \delta$ but $|f(x) - 1| \geq \epsilon$. This is the negation of the statement that $\lim_{x\to a} f(x) = 1$. That is, the limit cannot be 1.

score -1 · Answer 2 · answered Dec 23 '14 at 01:47

You do not simply "replace $x$ by the value $0$". What you do to arrive at the conclusion that $0$ is a limit (but $1$, $\pi$, whatever, are not), is realize that the condition "for $\varepsilon>0$, there exists $\delta>0$ ..." does not hold for $L=1$. For $L=1$ to be a limit, you must show that the condition holds for all $\varepsilon>0$, but you can show that it doesn't hold for $\varepsilon=1$, for example:

Let $\varepsilon=1$. Can I find a $\delta>0$ such that for all $x$ in the interval $D=\left(-\delta,\delta\right)\setminus\left\{0\right\}$ (that is, the interval without the zero) we have $\left|x-1\right|<\varepsilon=1$ ? No we cannot, because all the negative $x$'s have distances greater than $1$ to the supposed limit $L=1$.

P.S.: I am talking all the time about $0$ being "a limit" instead of "the limit" for $f\left(x\right)$ because the definition of "limit" states a condition that a number $L$ must hold in order to be called "a limit". However, you can show (as has been mentioned in the comments to your question) that if some number is "a limit", then it is the only limit.

Arriving at the value of Limits

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