Continuous function with $\delta - \epsilon$ I have one question to proof continuous function with $\delta -\epsilon$ definiton. I know how to prove, that function is continuous, but for some function we have to evaluate some term. For example: let $f(x):= x^2$, then we begin so:
Let $\delta>0$ and $|x-a|< \delta$, then $\forall \epsilon>0$ we have:
$|f(x)-f(a)|=|x^2-a^2|=|(x-a)(x+a)|$ and now we have a problem, that our term $(x+a)$ depends on x. So we have to evaluate it. We can say that $|x-a|<1= \delta$ or $|x-a|<2= \delta$ or every real positive number.
$\textbf{The question is:}$ Why do we have this posibilty to say it? Why can I say that $|x-a|<$ than every positive number? Why will this work?
Thank you for help
 A: First, I would not say that you have to evaluate the term $(x+a)$. But I would say that it is a good initial strategy to estimate that term, in particular to find some useful upper bound of the form $|x+a|< M$.
Remember, we are trying to prove the existence of an appropriate value of $\delta>0$, so our goal is to choose that value in some approriate manner and then to prove that $|x-a|<\delta$ implies $|x^2-a^2| < \epsilon$. On the way to choosing $\delta$, we are free choose an initial value $\delta_1$ and use the assumption $|x-a|<\delta_1$ to draw some conclusions, and then perhaps choose a secondary value $\delta_2 > 0$ and use the assumption $|x-a|<\delta_2$ to draw some conclusions, and perhaps even to chooose a tertiary value $\delta_3$ and so on. Once you have all your conclusions lined up, you can then set $\delta = \min\{\delta_1,\delta_2,\ldots\}$ and use the assumption $|x-a|<\delta$ to deduce that $|x-a|<\delta_1$ and that $|x-a|<\delta_2$ and so on, and then use the consequences of those deductions altogether to reach your final deduction of the inequality $|x^2-a^2|<\epsilon$.
We'll make our initial choice $\delta_1$ to let us get a useful estimate on $|x+a|$. Namely, let's simply take $\delta_1=1$. Assuming $|x-a|<\delta_1=1$, we conclude that
$$a-1 < x < a+1
$$
$$2a-1 < x+a < 2a+1
$$
and therefore
$$|x+a| < \max\{|2a-1|,|2a+1|\} = M
$$
$$|x^2-a^2| = |x-a| \cdot |x+a| < M |x-a|
$$
$$\frac{|x^2-a^2|}{M} < |x-a|
$$
And now we choose our a secondary value of $\delta$, namely $\delta_2 = \frac{\epsilon}{M}$.
And now we're ready to finish the proof: let $\delta = \min\{\delta_1,\delta_2\}$. Assuming that $|x-a|<\delta$ it follows that
$$\frac{|x^2-a^2|}{M} < |x-a|
$$
and therefore
$$\frac{|x^2-a^2|}{M} < \frac{\epsilon}{M}
$$
from which it follows that
$$|x^2-a^2| < \epsilon
$$
and the proof is done.
A: We don't say that $|x-a|$ is smaller that any positive number. We say that if $|x-a|<1$, then$$|x+a|=\bigl|(x-a)+2a\bigr|<1+2|a|,$$and that therefore$$|x^2-a^2|<\bigl(1+2|a|\bigr)|x-a|.$$So, taking $\delta=\min\left\{1,\frac\varepsilon{1+2|a|}\right\}$ will make us have$$|x-a|<\delta\implies\left|x^2-a^2\right|<\varepsilon.$$It turns out that the same argument works with any positive number. If I begin with, say $|x-a|<3$, then$$|x+a|=\bigl|(x-a)+2a\bigr|<3+2|a|,$$and therefore$$|x-a|<\min\left\{3,\frac\varepsilon{3+2|a|}\right\}\implies\left|x^2-a^2\right|<\varepsilon.$$
