Choosing a delta for the function f(x) = x^2

Question

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.

Answer 1

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 $$\delta=\min\left(r,\frac{\epsilon}{2|x_0|+r}\right).$$

Answer 2

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

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

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.

$|x-x_0|\le1\\ -1<x-x_0\le1\\ x_0-1\le x\le 1+x_0\\ 2x_0-1\le x+x_0\le 1+2x_0\\ |x+x_0|\le 1+2|x_0|$

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

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

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