How to determine the denominator/ nominator of a limit

[lilliv]

This is a problem and solution from my calculus course.

e = epsilon, d = delta

Prove directly using the formal definition: lim x-> 1 ([5x + 1]/[x^2 + x + 1]) < e

Solution:
abs([5x + 1] / [x^2 + x + 1] - 2) < e, we assume d < 0.1

abs([5x + 1] / [x^2 + x + 1] - 2)
= abs([5x + 1 - 2(x^2 + x + 1)] / [x^2 + x + 1])
= abs([2x^2 - 3x + 1] / [x^2 + x + 1]) * abs(x-1)
<abs([2]/[0^2 + 0 + 1]) * abs(x-1)
< 2d

I understand most of the solution except for the second last line, where values have been chosen for x (the upper bound and lower bound, I think). I know that the goal is to minimize the denominator and maximize the numerator. However, I am thoroughly confused as to how the upper and lower values for x are determined.

If anyone could explain how these x values are chosen, I would really appreciate it

 

[lippyka]

The upper and lower values for x chosen in this solution are determined by the delta, d. Since the goal is to find a range of x-values such that the absolute value of the expression is less than epsilon (e), the denominator must be minimized. The lower bound for x is determined by choosing the smallest possible value of x that is greater than 1-d. Similarly, the upper bound is determined by choosing the largest possible value of x that is less than 1+d. In this case, since d is set to 0.1, the lower bound for x is 1-0.1 = 0.9 and the upper bound is 1+0.1 = 1.1.