Understanding the Epsilon-Delta Definition of Limit in Calculus

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please guys! some one please help me understand the epsilon-delta definition of limit and the delta nbd concept.
 
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What have you been taught so far?
 
It's actually very simple. What does it intuitively mean when we say that \lim_{x\to p}f(x)=y? It means that if you let x approach p then f(x) will approach y. How to see if that is really the case? Well, if the function f is really going to get closer and closer to y if you let x appoach p closer and closer, then it should be the case that if you specify some small interval \delta containing y you can find an interval around p of length \epsilon such that the function f will map that interval within that deviation, no matter how small you make \delta.
 
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...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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