Is the Concept of Neighborhoods Essential for Understanding Limits in Calculus?

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Is it correct to think about the expresion:
"the limit of f(x) is b when x->a" as saying that for every x that's very close to a but not a (in the deleted neighborhood of a) there is a f(x) that's very close to b (in the neibourhood of b) - or is that not precise enough?
 
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It's definitely not precise enough for a mathematical definition. What do you mean by 'very close to a'. What is 'very close'?
The idea of \lim_{x\to a}f(x)=b is that you can make f(x) as close as you want to b by choosing x close enough (but not equal to) a. By close I mean that the distance |f(x)-b| can be made as small as we want. How small? Smaller than any given positive number.

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html
 
Might want to use 'gets closer' instead of 'is close to'

Thats pretty much the epsilon delta method.
 
Thanks guys for making that clear.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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