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I recently did all the limit maths, but I didn't put much thought into its significance. I thought, okay, it's perfectly reasonable to say that, for example, as $x\to 1$, $(\frac{x^2-1}{x-1})\to2$. However, more recently, I think I discovered that limits are far more useful than that. The approached value can be replaced with the indeterminate form itself!

For example, what's the slope of a point? Answer: $\frac{f(x+0)-f(x)}{0}=\frac{0}{0}$. We've reached an indeterminate form, so what? Just use limits: $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$. It's important to realise that the slope of a point is $\frac{f(x+0)-f(x)}{0}=\frac{0}{0}$, and by determining the slope of a point by $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, we are saying $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\frac{0}{0}$. The approached value can be replaced with the indeterminate form itself!...(i)

If that is the case, why don't we just define $\frac{0}{0}$ and other undefined/indeterminate forms then? Why keep this "undefined" thing in maths alive (I'm not saying that we actually do it; I just want to be sure that my reasoning is reasonable; that's why I'm asking this question)? The definition will be something like, "The value of an indeterminate form that is found by inputting a value in a function will be the limiting value of the function at that point."...(ii)

Questions:

  1. Am I right in (i) & (ii)? If I'm wrong in (i), how are mathematicians allowed to take the slope of the point as $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ when it should be $\frac{f(x+0)-f(x)}{0}$?
tryingtobeastoic
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    The problem with $0/0$ (and similar situations) is that you can find $f(x) \to 0$ and $g(x) \to 0$ where $f(x)/g(x)$ converges to any number you like. – Elchanan Solomon Oct 06 '21 at 11:42
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    $\frac{0}{0}$ itself is not defined because it could be any real (or complex) number, because $x\cdot 0=0$ holds for every $x$. In a limit, the variable APPROACHES the given value, it never IS the value. Nevertheless, the limit , if existent , is exact and not just an approximation. Concerning the slope : If $h=0$ , we have a single point and no slope. – Peter Oct 06 '21 at 11:42
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    Do you know the meaning of indeterminate quantities? Probably not since you are saying "why don't we just define 0/0 and other undefined/indeterminate forms then?" 0/0 can result different solutions in different cases. 0/0 can be anything. Have you solved such problems in which direct substitution results 0/0 form? Have you tried to know why does same 0/0, yields different solutions in different cases? How can you define 0/0? –  Oct 06 '21 at 11:45
  • @Prothala Yes, I understand that the value of $\frac{0}{0}$ can be anything. My beta definition allows for $\frac{0}{0}$ to be anything: "The value of an undefined form that is found by inputting a value in a function will be the limiting value of the function at that point." – tryingtobeastoic Oct 06 '21 at 12:23
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    @tryingtobeastoic Do you know that there's a difference between indeterminate quantites and undefined quantities? There are basically $7$ Indeterminate quantities whereas any fraction where denominator is $0$ is undefined. I've edited your question to fix this mistake. –  Oct 06 '21 at 16:37
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    The approached value can be replaced with the indeterminate form itself!. Your this statement seems to be incorrect. Let's consider the example mentioned by you in the question $\lim\limits_{x\to 1}(\frac{x^2-1}{x-1}) = 2$ Here your statement is that the approached value which is $2$ can be replaced with the indeterminate form which is $0/0$. Do you think that this even make any sense? Inorder to avoid these indeterminate quantities, we introduced the concept of limits. We cannot replace 2 with 0/0. That's all doesn't make any sense I think. –  Oct 06 '21 at 16:44
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    By determining the slope of a point by $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, we are saying $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\frac{0}{0}$. Your this statement which is used in the first example is also wrong. How can we say $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\frac{0}{0}$? We have to avoid these indeterminate quantities by the concept of limit. You cannot actually compare the limiting value of a function at a certain point say $a$ with the value of the function at $a$ which doesn't even exists. –  Oct 06 '21 at 16:51

1 Answers1

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how are mathematicians allowed to take the slope of the point as $\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ when it should be $\displaystyle\frac{f(x+0)-f(x)}{0}$?

$\displaystyle\frac{f(x+h)-f(x)}{h},$ called the difference quotient, is just a straight line's slope, which in turn is the vertical change divided by the horizontal change, in comparing two distinct points on the straight line. Thus, $h$ is never zero; it simply isn't meaningful.

The point is this: when computing $f'(c),$ we are taking infinitely many slopes that pivot about the reference point $c;$ this ‘pivoting’ stabilises as the comparison points approach $c;$ as the slopes correspondingly stabilise, the ideal best value is assigned to $f'(c).$

ryang
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    please explain the solutions of all the questions which are asked by questioner. –  Oct 07 '21 at 15:36