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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?

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  • $\begingroup$ For your choice $\lim_{x\to c}x=c$ $\endgroup$ – janmarqz Dec 23 '14 at 0:59
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    $\begingroup$ 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 $c=c'$. $\endgroup$ – Git Gud Dec 23 '14 at 1:09
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    $\begingroup$ Note that the ability to replace the limit at $x$ with the value $f(x)$ means that the function is continuous at $x$. $\endgroup$ – Mark Bennet Dec 24 '14 at 7:28
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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.

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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.

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