CONTINUITY In an informal way, we can say that a function f is continuous on an interval if its graph can be drawn without taking the pencil off of the paper ( f doesn’t cut or jump). More precisely f is continuous at x = c if and only if: I) f(c) is defined, and II) lim f(x) = f(c) . x c (meaning you can get to the point (c,f(c) ) by walking on the curve from left or from the right of it) A function f is said to be continuous on an interval I (domain) if and only if f is continuous at any point in I. Note that if f is continuous on I, then lim f(x) coincides with f(c) x c Ex 1. Find where f(x) = (2x-4) / (x2 –x –6) is not continuous ( points of discontinuity ). Since f(x) is not defined at x where its denominator is zero. The discontinuity of f occurs when x2 –x –6 = 0. Solving x2 –x –6 = 0, we get x=-2 or x= 3. So the values for which f is discontinuous are x=-2 or x=3. Ex 2. The line f(x) = ax+b is continuous in its domain R. f(c) = ac+b exists and lim . . x c (ax+b) =ac+b = f( c ), so f is continuous at any real c Ex 3. Polynomials p(x) are continuous on R, since lim p(x) = p(a) and p(x)| x=a= p(a) . xa

C0NTINUOUS FUNTIONS V/S DISCONTINUOUS FUNCTIONS Analyze continuity for each of the four functions illustrated with the graphs shown below. 1) y = f(x) is continuous, since f has no disconnecting places or jumps. Y y=h(x) y= f(x) 2) y = g(x) is continuous, since g has no disconnecting places or jumps. y= g(x) y=r(x) 3) y = h(x) is discontinuous , since h has a jump at x = a. Jump X 4) y = r(x) is discontinuous, since r is disconnected at x=c. c There’s a hole. A point is missing!

MORE PROPERTIES OF CONTINUOUS FUNCTIONS Theorem: (Combination of continuous functions). If f and g are continuous at x=c then the following combinations of them are continuous at x=c too. 1) Scalar multiple: kf is continuous, for any constant k 2) Sum and Difference: f +g and f - g are continuous 3) Product: fg is continuous 4) Quotient: f/g is continuous if g(c) ≠ 0 y = 3x 5) Composition: fog is continuous . (provides that f is continuous at g(c) ) y = x2 ` Y Example 1: We know f(x) = x is continuous. The following functions are continuous too. See figure on the right. y=x2 - 4x+ 5 y = x y=x-1 1) 3f(x) = 3x 2) ( ff)(x) = f(x)f(x) = xx = x2 X 3) g((x) =( f•f)(x)+4 f(x)+5 = x2 - 4x + 5 2 2 2 4) h(x) = (x2 - 3x + 4 ) / (x-2) Note that x =2 is a Vertical Asymptote for h(x) So h(x) is continuous for any x  2 . And the oblique asymptote is y=x-1.

The above examples show us that polynomial functions are continuous functions on R and that the rational functions q(x) = p1(x) / p2(x) (where p1 & p2 are polynomials) are continuous on R – { the zeros of p2(x) }. Example 1: Study continuity for f(x) = (x+1)/ (x2 – x –12) We look for the zeros of denominator, i.e. x2–x–12 =0 (x +3)(x – 4)=0. So x=4 or x= - 3. So f(x) is continuous on R – {-3,4} Example 2: Study continuity for g(x) = sin2 x /(x4 + 1). (Assume that sinx is continuous) Since sin x is continuous its, square is also continuous. Now not only that the denominator is a continuous function (it’s a polynomial ! ). Furthermore x 4 + 1 cannot be zero, since x4 + 1 is always  1 so . So g(x) is continuous on R. Example 3: Study continuity for sin x & cos x Since sin(x+h) = sin x cos h + cos x sinh we have 1 lim sin(x+h) = lim (sin x cos h + cos x sinh) . h 0 . h0 = sin x lim cos h + cos x lim sin h . . h 0 h0 But we know that lim sin = 0 & lim cos h = 1 . h0 h0 So lim sin(x+h) = sin x and this means that sin x is a continuous function. . h 0 Similarly cos x is a continuous function using cos(x+h) = cos x cos h - sinx sinh

Intermediate Value Theorem Theorem: If f(x) is a continuous function on a closed interval [a,b] , then f(x) must take all values between f(a) and f(b). Y f(a) f(b) The figure on the right shows a continuous function on [a,b] c The Theorem assures us that if we choose any number c between f(a) and f(b) , the line y=c will have to intersect at least one point on the graph. In our case it cuts the graph at three points (x1, c) , (x2 , c) & (x3 , c). X a b x1 x2 x3 Example: Prove that f(x) = x3+5x – 7 has a zero between 1 and 2. Y (2,7) 7 We know that f(x) = x3+5x – 7 is a continuous function. X Since f(1) = 1+5- 11 = - 5 & f(2) = 8+10-11= 7. We have f(1)<0<f(2). - 5 (1,-5) The Intermediate Theorem assures us that the line y=0 intersects the graph of f(x) at a value of x between 1 and 2.

Example 1: Find lim f(x) , . x 16 Theorem: If f is continuous at g(x) then lim f( g(x) ) = f( lim (g(x) ) . . . xa xa Example 1: Find lim f(x) , . x 16 for f(x) = x3/2 . (Assume that f(x) is continuous) Interchanging lim and f we get lim f(x) = [ lim x ]3/2 =163/2 = 64 x 16 x 16 Example 2: Find lim . x  ln [(3x+1)/(x+9)] (Assume that ln x is continuous for x>0) Interchanging lim and ln, we get ln [ lim ((3x+1)/(x+9))] . x  = ln(3) Example 3: Find lim f(x) , . x-1 for f(x) = (sin [(x2 – 1)/(x2+4x+3) ] )2 Interchanging lim and ( )2 …. {lim sin[(x2 – 1)/(x2+4x+3) }2. . x -1 Interchanging lim and sin ….. {sin lim (x2 – 1)/(x2+4x+3) }2. . x-1 But lim (x2 – 1)/(x2+4x+3) is type 0/0 , so we try to simplify by (x+1) …. . x-1 2 2 Factoring we get lim (x2 – 1)/(x2+4x+3) = lim (x+1)(x-1) / (x+1)(x+3) . x -1 x -1 = - So lim f(x) = (sin (-))2 = 0 . x-1