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I have just started with the epsilon-delta definition of limits. For the function f(x) = $x^2$, I do not understand why the following delta was chosen, how it was derived and why it is the right one. $δ=\min\left(1,\frac{ϵ}{2|x_0|+1}\right)$

My mathematical maturity is at a very primitive level and I would be very grateful if you can give a very detailed explanation that doesn't assume anything more than algebra and the basic definition of a function approaching a limit. That's all I know. Verbose answers are also welcome.

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    What is $x_0$ and what are you trying to show using epsilons and deltas? – 2017-02-27
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    I can assure you that everyone knows what I'm talking about if they simply have gone through a few pages of a Calculus book. As ambiguous as my question is, the number of the times the function has been used as an example makes it very close to most people. – 2017-02-27
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    I guess we assume you are trying to prove continuity then. – 2017-02-27
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    After checking your account I think you have gone so forward in Mathematics that you have forgotten this example in your first real analysis book back in the days. ;) The question is about proving that the function approaches the limit $x_o^2$ at $x_o$ – 2017-02-27

2 Answers 2

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Let $\epsilon>0$ and assume that $0<\delta\leq 1$. Note that if $|x-x_0|<\delta$ then, by the triangle inequality, $|x|\leq |x_0|+|x-x_0|< |x_0|+\delta\leq |x_0|+1$. Moreover $$|f(x)-f(x_0)|=|x^2-x_0^2|=|x+x_0||x-x_0|\leq |x+x_0|\delta\leq (|x|+|x_0|)\delta< (2|x_0|+1)\delta.$$ How can you define $\delta>0$ in such a way that $(2|x_0|+1)\delta\leq \epsilon$?

P.S. Note that the bound $1$ is arbitrary. We can also replace it with any positive constant $r$. Then we define $$\delta=\min\left(r,\frac{\epsilon}{2|x_0|+r}\right).$$

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$\lim_\limits{x\to x_0} x^2 = x_0^2$

$\forall \epsilon > 0, \exists \delta >0 : |x-x_0|<\delta \implies |x^2 - x_0^2| < \epsilon$

If we can find a value of $\delta$ then we will have shown that the limit exists and equals what we need it to equal.

$|x^2 - x_0^2| = |x-x_0||x+x_0| < |x+x_0|\delta \le \epsilon$

We would like to define $\delta$ in such a way that does not depend on $x.$

Choose a boundary for $\delta$. In this case, we have a lot of room to choose such a boundary. $1$ is always an easy number to work with. It is somewhat arbitrary. But, nothing says we can't choose 1.

$|x-x_0|\le1\\ -1

$(1+2|x_0|)\delta \le \epsilon$

$\delta \le \frac {\epsilon}{1+2|x_0|}$ and $\delta \le 1$

$\delta = \min (1, \frac {\epsilon}{1+2|x_0|})$