Quadratic functions

Now look at f(x) = x². We want to say $\lim_{x\rightarrow 11} f(x) = 11^2$. So for every t > 0 we have to produce m_t > 0 so that if 0 < |x-11| < m_t then |f(x) - 11² | < t.

$$\vert f(x) -11^2\vert = \vert x^2 - 11^2\vert = \vert(x-11)(x+11)\vert = \vert x+11\vert\cdot\vert x-11\vert$$

This is a whole new ballgame. Happily we see that we have a factor |x-11| but it is multiplied by a factor which depends on x. Let us make a preliminary stab and require that 0 < |x-11| < 1. Then |x+11| < 23 and we have

$$\vert f(x) -11^2\vert < 23\vert x-11\vert < 23 \;\; {\rm if}\;\;0 < \vert x-11\vert < 1.$$

Well, if $t \geq 23$ we are all set. But what if it isn't? Our experience with 2x suggests we just reduce t by whatever multiplies x, in this case 23, suggesting we require 0 < |x-11| < t/23. To cover both possibilities, we simply take $m_t = \min(1,t/23)$.