Do now as a warm up: Is there some number a, such that this limit exists? If so, find the value of a and find the limit. If not, explain why not. 1
Continuity and One Sided Limits To say that a function is continuous at x = c means that there is no interruption in the graph of f at c. From a visual perspective, a continuous function is simply a nice, smooth function that can be drawn completely without lifting your pencil. The graph is unbroken at c There are no holes, jumps, or gaps. Whereas, limits told us where a function intended to go, continuity guarantees that the function actually made it there. 2
Three conditions exist for which the graph of f is not continuous at x = c. exists f(c) exists A function is not continuous at x =c if: 1. function is not defined at x = c 2. limit of f (x) does not exist at x = c 3. limit of f (x) exists, but it is not equal to f (c) 3
Definition of Continuity A function f (x) is continuous at a point x =c if three conditions are satisfied: 1. exists 2. is defined 3. test this first means function has an intended height means there is no hole there means function's value matched its intended value 4
EX #1: Determine whether or not the function is continuous at x = 1. First Test g(x) is not continuous at x = 1 because the limit as x approaches 1 is not equal to the function value at 1. 5
LIMITS 6
Definition of Continuity on a Closed Interval A function f is continuous on a closed interval [a, b] if it is continuous on the open interval (a, b) and and 7
EX #2: Discuss the continuity of the function The domain of f is all real numbers except x = 2. You can conclude that f is continuous at every x value in its domain. At x = 2, the function has a removable discontinuity. If f(2) is defined as 1, the "newly defined" function is continuous for all real numbers. 8
EX #2: Show that f is continuous on [2, 6] given 9
EX. #3: Discuss the continuity of the composite function f(g(x)) if... and At x 1, the function is discontinuous. However, if f(g(x)) is defined as x >1, the "newly defined" function is continuous at every x value in its domain. and The domain of f(g(x)) is all real numbers except x = 6. At x = 6, the function is discontinuous. You can conclude that f(g(x)) is continuous at every x value in its domain. At x = 6, the function has a non removable discontinuity since the limit as x 6 does not exist. 10
Properties of Limits If b is a real number and f and g are continuous at x = c, then the following functions are also continuous at c. LIMITS CONTINUITY bf f ± g fg f/g ; g 0 Polynomial, Rational, Radical, Trigonometric functions are continuous at every point in their domains. 11
Types of Discontinuity Jump Discontinuity typically caused by a piece wise defined function Point Discontinuity occurs when a function contains a hole Infinite Discontinuity occurs primarily at a vertical asymptote Summary: 1. In no limit exists, you have jump discontinuity. 2. If the limit exists but the function doesn't, you have point discontinuity. 3. If the limit doesn't exist because it is or, then you have infinite discontinuity. 12
Pull Pull Pull Pull Pull Classifying Discontinuities REMOVABLE NON REMOVABLE POINT (Holes) 2 sided limit exists JUMP 1 sided limits exist INFINITE (vertical asymptotes) at least one of the 1 sided limits doesn't exist Pull 13
EX #4: Find the domain for each function. Where is each function not continuous? What kind of discontinuity is this? X<2 non-removable X=-6, x=1 non-removable 14
Intermediate Value Theorem A continuous function on an interval cannot skip values. An important outcome of the I.V.T. is that it can be helpful in finding zeros of a continuous function on an interval. 15
Thm. The Intermediate Value Theorem (IVT) f is continuous on [a,b] m R f(a)<m<f(b) Try it! Draw a cont's function between (a,f(a)) and (b,f(b)). Pick any y value between f(a) and f(b). A horizontal line through that y value has to cross your function. http://www.calculusapplets.com/ivt.html a [a,b] where f(c)=m This is an existence theorem. It says something exists, but not where. f(a) m f(b) c b 16
The IVT holds for these: f is continuous on [a,b] m R f(a)<m<f(b) The IVT even holds for this: It says there's at least one c value that works. There could be many! But, if f is not continuous on [a,b], the theorem doesn't tell us anything. 17
EX #5: Apply the IVT, if possible, on [0,5] so that f(c) = 11 for the function 1. f is a polymomial, so it is continuous everywhere. 2. 3. -1 29 4. Therefore, on the interval [0,5] f(c)=11 when x =3. 18
Apply the IVT, if possible, on so that f(c) = 0. for 1. f is a trigonometric, so it is continuous everywhere. 2. 3. 0-1 and 19