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Is not calculus based on the paradox that the closest point to a point A is a distinct point B which is the point A itself?

For example, if we consider the limit,

$$ \lim_{x\to2} \frac{x^2-4}{x-2} $$

It's evaluated by first cancelling out the $(x-2)$ common factor and then we substitute the value $2$ in the function of $x$. It's like, at first, we're considering that $x$ is nearly equal to $2$ but not $2$, but then we substitute the value of $2$. So, what's happening here? Similarly, in derivatives we're considering the tangent to a curve which intersects the curve at one distinct point. That's how we get the exact slope. But, at times, we're considering a point $A$ which is close to point $B$ (and point $B$ and point $A$ are different). But, we know that the concept of derivatives is legit by experimental evidence. So, how does it all actually work?

Anubhab Das
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    You are interested in learning more about epsilon-delta proofs; you might find an example like this one to be helpful in understanding what limits actually mean. – Clayton Aug 22 '15 at 14:27
  • But how can it work if it has two opposite implications? – Anubhab Das Aug 22 '15 at 14:28
  • What do you mean by "two opposite implications" in this context? – Clayton Aug 22 '15 at 14:29
  • At one point, we're substituting 0, but before that we're cancelling out factors based on our assumption that the function tends to 0. We can't simultaneously perform both operations as they're kinda contradicting, right? Substituting 0 in this context makes our prior assumption invalid. – Anubhab Das Aug 22 '15 at 14:34
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    Note that you don't need to cancel common factors in your example. You can substitute $x=0$ to get $\frac{-4}{-2}=2$. (It would be different if we were taking the limit as $x \to 2$.) – Théophile Aug 22 '15 at 14:39
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    If you were to draw the graph it would be $x-2$ but with the point $x=2$ missing. So when you do the limit it's trying to figure out where the graph would be at $x=2$. – Dr Xorile Aug 22 '15 at 14:40
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    @AnubhabDas It does not matter if you sub $x=0$ in the given fraction or in $x+2$ because for all $x$ not 2, these expressions are equal to each other. You are not canceling factors based on the assumption that the function tends to go to zero. You are canceling factors because the algebra allows you to do so under the condition that you knock out $x=2$ from the domain, which is why we have limits to see what is going on around $2$ – imranfat Aug 22 '15 at 14:41
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    One must suspect that $x$ approaching $2$, so that $x-2$ approaches $0$, was intended. ${}\qquad{}$ – Michael Hardy Aug 22 '15 at 14:42
  • possible duplicate of What exactly is going on when we're finding a limit? –  Aug 22 '15 at 20:55
  • I don't think this question should have been closed. Sometimes a question can amount to "How should I make this idea precise?" and from one point of view it's unclear what is being asked, but from the point of view appropriate to the occasion, it's clear. ${}\qquad{}$ – Michael Hardy Aug 22 '15 at 23:07
  • I think the root cause of such confusion is the way limits are introduced to students in non-rigorous fashion. It is better to understand the definition of limit properly and then one finds that there is no mention of "substituting the value of $x$" in the definition to find the limit (in fact definition does not give a way to find limit, it just proposes a way to check if a given number is a limit or not). So calculus is not based on paradox, rather the paradox is created by people (mainly instructors and crappy book authors) who want to make calculus "too intuitive to be true". – Paramanand Singh Aug 23 '15 at 04:39
  • A limit is not evaluated by substituting the value of $x$. Please see the answer http://math.stackexchange.com/a/1399974/72031 which elaborates more on this point. – Paramanand Singh Aug 23 '15 at 04:47

2 Answers2

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What is happening is that the two functions $$ x\mapsto \frac{x^2-4}{x-2} \text{ and } x\mapsto x+2 $$ both have the same values at all inputs $x$ except $x=2$, and the first function is undefined at $x=2$ and the second is not only defined but continuous at $x=2$.

To say that the limit is $4$ means that $f(x) = \dfrac{x^2-4}{x-2}$ can be made as close to $4$ as desired by making $x$ close enough, but not equal, to $2$.

Suppose you want to make $f(x)$ between $4-0.0000001$ and $4+0.0000001$. There is some number $\delta$ so small that if $x$ is between $2-\delta$ and $2+\delta$ but not exactly $2$, then $f(x)$ is between those bounds. And if you want $f(x)$ to be between $4-0.0000000000001$ and $4+0.0000000000001$, you can guarantee that by making $\delta$ smaller.

Michael Hardy
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Until the early 1800s, you would have been essentially correct in your understanding. Calculus was considered an intuitive technique rather than a rigorous mathematical theory. Cauchy changed that by establishing a rigorous theory for it.

Paul Sinclair
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