By Gill G.S.
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Example text
19 Suppose that f (x) ≤ g(x) ≤ h(x) for all x in an open interval containing c and lim f (x) = lim h(x) = L. x→c x→c Then, show that, lim g(x) = L. x→c Let > 0 be given. Then there exist δ1 > 0, δ2 > 0, and δ = min{δ1 , δ2 } such that |f (x) − L| < |h(x) − L) < 2 2 whenever 0 < |x − c| < δ1 whenever 0 < |x − c| < δ2 . If 0 < |x − c| < δ1 , then 0 < |x − c| < δ1 , 0 < |x − c| < δ2 and, hence, − < f (x) − L < g(x) − L < h(x) − L < . 2 2 It follows that |g(x) − L| < 2 < whenever 0 < |x − c| < δ, and lim g(x) = L.
A vertical line has no slope. A horizontal line has an equation of the form y = c. A horizontal line has slope zero. A line that is neither horizontal nor vertical is called an oblique line. 62 CHAPTER 2. LIMITS AND CONTINUITY Suppose that an oblique line passes through two points, say (x1 , y1 ) and (x2 , y2 ). Then the slope of this line is define as m= y2 − y1 y1 − y2 = . x2 − x1 x1 − x2 If (x, y) is any arbitrary point on the above oblique line, then m= y − y2 y − y1 = . x − x1 x − x2 By equating the two forms of the slope m we get an equation of the line: y − y1 y2 − y1 = x − x1 x2 − x1 or y − y2 y2 − y1 = .
13 Show that every polynomial P (x) is continuous at every c. From algebra, we recall that, by the Remainder Theorem, P (x) = (x − c)Q(x) + P (c). Thus, |P (x) − P (c)| = |x − c||Q(x)| where Q(x) is a polynomial of degree one less than the degree of P (x). As in Example 12, |Q(x)| is bounded on the closed interval [c − 1, c + 1]. For example, if Q(x) = q0 xn−1 + q1 xn−2 + · · · + qn−2 x + qn−1 |Q(x)| ≤ |q0 | |x|n−1 + |q1 | |x|n−2 + · · · + |qn−2 | |x| + |qn−1 |. Let m = max{|x| : c − 1 ≤ x ≤ c + 1}.