Thank you both. @Math_QED and @PeroK
Well, I am still learning, and when I come upon stuff like this, I cannot help but feel frustrated for not understanding. And I don't really trust my knowledge enough at this point to judge by myself that such and such is a typo/mistake or not.
Yes, I do. ##|f(x)-f(x_0)|=5|x-x_0|<\epsilon##, and so we have ##\delta=\epsilon/5##.
But this doesn't show that ##|x-x_0|\leq|f(x)-f(x_0)|/5## for ##|x-x_0|<\delta##, which is what I am asking about. I really do understand the continuity proof, trust me!
@PeroK @Math_QED no, I don't want to prove ##f##'s continuity, rather I wanted to verify the author's claim that:
I can see that ##|x-x_0|=|y-y_0|/5##, and this makes the phrase "is sufficiently small if it does not exceed" a bit confusing. The equality always hold, no matter how small is...
I am reading from Courant's book. He gave an example of the continuity of ##f(x)=5x+3## by finding ##\delta=\epsilon/5##. He then said that ##|x-x_0|## does not exceed ##|y-y_0|/5##, but I don't see how he came up with this inequality.
I know that ##|x-x_0|<\epsilon/5##, and that...
In chapter 1, page 10, real numbers are found by confining them to an interval that shrinks to "zero" length (we consider subintervals ##I_0,\,I_1,...,\,I_n##). Basically, if ##x## is between ##c## and ##c+1##, then we can divide that interval into ten subintervals, and we can, then, have...