limit &continuity

CALCULUS 2010-2011 2010-2011 CHAPTER 1 Square: Area = L2 Perimeter = 4L Diagonal = √2L Rectangle: Area= LW Perimeter = 2 (L+w) Triangle: Area = 1/2 bh = 1/2 × side × side × sine included angle Perimeter = sum of lengths of the sides Equilateral triangle: Area= Perimeter =4x Trapezium: Area= 1/2 (b1 + b2 ) × h Isoscales trapezium: b1 = b2 +2 x Rhombus : Area = Perimeter = 4x Circle : Area = (r2 Circumference = 2( r Prism: Volume = area of base × height Lateral Area = perimeter of base × height Total surface area = area of base +lateral area Cuboid : V = Lwh L.A = 2 (L+w) h B.A = 2Lw T.S.A = 2(Lw + Lh + wh) Cube: V = X3 L.A= 4X2 B.A.= 2X2 T.S.A= 6x2 Cylinder: V= ( r 2 h L.A = 2 (rh B.A= 2( r2 T.S.A = 2(r (r + h) Sphere : V= T.S.A = 4( r2 Limits Revision Limit of a function defined by more than one rule Definition Solved Examples 1 Solution 2 Find Solution 3 Solution 4 Solution 5 Solution 6 Solution f (x) exists -5 = -a +b (1) f (x) exists 3a +b =3 (2) (2) – (1) 4a = 8 (a = 2 Solve in (1) (b = -3 Continuity of a Real Function at a point F(x) is continuous at x = a if : 1) f (x) is defined at X = a Solved Examples Discuss the continuity of the following function at the indicated points: ( f (x) is cont. at x = 0 (f(x) is cont. at x = 2 (f(x) is cont. at x = 0 (f (x) is discont. at x = 0 (f (x) is discont. at x = 0 Redefine the following functions such that each becomes cont. at the indicated pt. b = 3 Continuity of a Function on an interval 1) f (x) is continuous on ] a , b [ If f (x) is continuous at every point x ( ] a , b [ 2) f (x) is continuous on [ a , b ] if : a) f (x) is cont. on ]a, b[ b) f (x) is cont. at x = a, x = b Note: polynomial functions are continuous on R Rational functions f (x) = are continuous on R – Z {R(x)} where Z {R(x)} is the set of zeros of R (x) Trigonometric functions a. F (x) = sin x , f (x) = cos x continuous on R b. f (x) = tan x is continuous on R - solved Examples Discuss the continuity of each Function F (x) on R: 1 F(x) = x2 + 5x +1 f(x)is polynomial function ∴f(x)is cont. on R 2 f (x) = X2 – x – 20 = 0 (x – 5 ) ( x + 4 ) = 0 x = 5 , x= -4 Cont. on R – {5 , -4} F (x) is polynomial function F (x) is cont. on these intervals at x = -1 f (-1) = -1 , f ( -1-) = -1 , f(-1+) -3 + 2 =-1 ( f (x) is cont . at x = -1 at x = 3 f (3) =2(9)-7=11 =f (3-) = f(3+) ( f (x) is cont . at x = 3 ( f ( x) is cont. on R 6 Find the value of a such that f (x) is continuous: a Kinds of the roots 2nd degree equation Solved Examples: ax2 + bx + c = 0 45 1 1 w 1 h b b1 h b2 b2 b2 b1 x x x x x x r h w I x x x r h F (x) = 2/3(7-4x) F (x) = 2 -222222 F (x) = F(x)= F(x)=-1 3x+2 -1 3 2x2 – 7 -(/4 Sin x cos x 2(