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Let a < c < b and let f be a function whose domain contains (a, c) (c, b) We say that f has limit at c, and we write limx c f (x) = when this condition holds: For each > 0 there is a > 0 such that | f (x) | < whenever 0 < |x c| < It is important to know that there is a rigorous definition of the limit concept, and any development of mathematical theory relies in an essential way on this rigorous definition However, in the present book we may make good use of an intuitive understanding of limit We now develop that understanding with some carefully chosen examples

If you have trouble seeing why the above equation is in polynomial standard form, you can rename the coefficients and constants like this: pr = a ps + qr = b qs = c By substitution, you get ax 2 + bx + c = 0 If you wonder about the requirement that p 0 and r 0 in the binomial factor form, the reason for this restriction should be clear now If p = 0 or r = 0, then pr = 0, the term containing x 2 vanishes, and you have the one-variable first-degree equation (ps + qr)x + qs = 0

What are the roots If we find a quadratic equation in binomial factor form and we want to find the roots, we re in luck! Here is the binomial factor form again, for a quadratic in the variable x :

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1 Gary H Anthes, Moniker Mania Computerworld (May 18, 1998): 65 2 Leslie Goff, The Skills That Thrill, Computerworld (December 4, 2000): 57 3 Ibid

(px + q)(rx + s) = 0

See Figure 21 Calculate limx 1 f ( x)

where neither p nor r equals 0 If either factor (that is, either binomial) happens to equal 0, then the entire expression on the left side of the equals sign becomes 0 The equation then reduces to 0 = 0, indicating a root! The roots of the above quadratic can therefore be found by solving these two first-degree equations: px + q = 0 and rx + s = 0 In the first equation, we can subtract q from each side, getting px = q Dividing through by p, which we have said is nonzero, we get x = q /p In the second equation, we can subtract s from each side to obtain rx = s Then we divide through by r, which we have restricted to nonzero values, getting x = s /r The roots of the quadratic are x = q /p or x = s /r If X is the set of solutions, called the solution set, then X = { q /p, s /r} The solution set for an equation with multiple roots is the set containing all of those roots As you continue studying algebra, you ll come across equations with solution sets containing three, four, or more elements The solution set for a true quadratic, however, never has more than two elements

Here s a challenge!

Observe that, when x is to the left of 1 and very near to 1 then f ( x) = 3 x is very near to 2 Likewise, when x is to the right of 1 and very near to 1 then f ( x) = x 2 + 1 is very near to 2 We conclude that lim f ( x) = 2

Consider the following quadratic equation in standard form: x 2 2x 15 = 0 Put this equation into binomial factor form Then find the solution set X Here s a hint: The coefficients and constants in the binomial factor form are integers

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