Schaum's Outline of Theory and Problems of Beginning Calculus

CHAP. 83 LIMITS 63(b) f(x + h) = 4(x + h)’ - (X + h) = 4(x2 + 2hx + h2) - x - h= 4x2 + 8hx + 4h2 - x - hf(x) = 4x2 - xf(x + h) -f(x) = (4x2 + 8hx + 4h2 - x - h) - (4x2 - X)= 4x2 + 8hx + 4h2 - x - h - 4x2 + x= 8hx + 4h2 - h = h (8~ + 4h - 1)f(x + h) -f(x) h(8x + 4h - 1)h- =8~+4h- 1h1(c) f(x + h) = -x+h= lim (8x - 1) + lim 4h = 8x - 1 + 0 = 8x - 1h-0 h+O1 1x+h xf(x + h) -f(x) = - - -ALGEBRA---=ac ad - bcb d bd- c- ---(x + h)x (x + h)x (x + h)xx-(x+~) x - x - ~ -hHence,and.-= --.- =--1 1 1 1 1=- lim (x + h) x x x X2h+Ox3 - 18.3 Find lim -x-1 x - 1 -Both the numerator and the denominator approach 0 as x approaches 1. However, since 1 is a root ofx3 - 1, x - 1 is a factor of x3 - 1 (Theorem 7.2). Division of x3 - 1 by x - 1 produces the factorizationx3 - 1 = (x - 1Xx2 + x + 1). Hence,x3 - 1 (x - lXX2 + x + 1)lim - = lim=lim(~~+x+1)=1~+1+1=3x+l x- 1 x-1 x-1 X+ 1[See example (b) following Property V in Section 8.2.18.4 (a) Give a precise definition of the limit concept; lim f(x) = L.x+a(b) Using the definition in part (a) prove Property V of limits:limf(x) = L and lirn g(x) = K imply lim [f(x) + g(x)] = L + Kx+a x-a x-a