There is probably a way to
write the proof using mostly references to absoulte values. However, it is useful to know how to "grunge it out" when no elegant way comes to mind.
When you have to get down and dirty, it is best to write things like |y-7| < \delta in the equivalent form of:
eq 1. 7 - \delta < y < 7 + \delta
(For simplicity I'll label them "equations" but the they actually are inequalities.)
To square eq. 1 and keep the inequality marks pointed the same way, we must make sure that all the terms are positive. We can make 7 - \delta > 0 by chosing \delta < 7, so remember this condition. Squaring eq 1., we get:
eq. 2. 49 - 14 \delta + \delta^2 < y^2 < 49 + 14\delta + \delta^2
To get the functions of \delta to be closer to y^2 than \epsilon we need eq. 3 and eq. 4 to hold:
eq 3. 49 - \epsilon < 49 - 14 \delta + \delta^2
eq. 4. 49 + 14\delta + \delta^2 < 49 + \epsilon
Thos equations simplify to eq 5. and eq 6. respectively:
eq 5. -\epsilon < -14 \delta + \delta^2
eq 6. 14 \delta +\delta^2 < \epsilon
Mutliplying eq 5. by -1 and reversing the inequality sign gives:
eq 7. 14 \delta - \delta^2 < \epsilon
If eq. 6 holds then eq 7 would also, so we only worry about eq 6.
Rather than worry about solving quadratic equations, it's simpler to take advantage of the fact that we are dealing with inequalities and trying to make \delta small.
So add the condition 0 < \delta < 1 so that we can say \delta^2 < \delta
This and eq 6. imply that we want:
eq 8. 0 < 14 \delta + \delta^2 < 14\delta + \delta < \epsilon
eq 9. 15 \delta < \epsilon
So this imples we want:
eq 10. \delta < \frac {\epsilon}{15}
We can satisfy eq 10. by setting \delta equal to various things, for example \delta = (0.5)\frac{\epsilon}{15} or \delta = \frac{\epsilon}{16} etc.
We have to remember the previous assumptions we made on \delta.
To incorporate all of them , it is sufficient to say:
eq 11. Let \delta = min\{ \frac{\epsilon}{16}, 1.0 \}
To have a real proof you have to go through the reasoning in reverse order.