Presentation on theme: "One-Sided Limits and Continuity"— Presentation transcript:
1 One-Sided Limits and Continuity ByDr. Julia Arnold
2 One-Sided LimitsThe function f has the right-hand limit L as x approaches a from the right writtenIf the values f(x) can be made as close to L as we please by taking x sufficiently close to (but not equal to) a and to the right of a.The function f has the left-hand limit M as x approaches a from the left writtenIf the values f(x) can be made as close to M as we please by taking x sufficiently close to (but not equal to) a and to the left of a.
3 Theorem 3: Le f be a function that is defined for all values of x close to x = a with the possible exception of a itself. ThenIf and only ifThus the two-sided limit exists if and only if the one-sided limits exist and are equal.Example 1: LetSince = -0 =0 both the left and right limits exist and are equal thus the limit is0.Find
4 Example 2: LetFindTheIs the right hand limit.Is the left hand limitSince they are unequal the limit of g(x) as x approaches 0 does not exist.
5 Continuous FunctionsA function f is continuous at the point x = a if the following conditions are satisfied.1. f(a) is defined.exists3.If a function is not continuous at a point then it is discontinuous.
6 ABOn the graph we can see that there are 2 points of discontinuity.Let’s see which of the three properties are violated.Is f(A) defined? Yes (the solid dot)Does the exist? No the right limit and left limit are not the same.Is f(B) defined? No
7 We will now look at 3 functions all of which are discontinuous at some point. We will also examine which of the 3 properties is violated.Since f(x)= x+2 is a straight line and straight lines are continuous, we can conclude that the discontinuity must come from the piecewise definition of the new function and the discontinuity must occur at x = 1.By definition f(1) = 1 but on the line f(1) = 3 which implies thatEquation 1:F(x) is discontinuous at x=1because it violates the 3rd propertyHence, property 1 and 2 are okay.Property 3 is violated which requires that
8 Equation 2:SinceWe can conclude that this is a straight line x+2 but we know that x because of division by 0 thus it is a straight line with a hole in it at x =2. The discontinuity occurs at the domain problem x = 2 and is discontinuous because f(2) is not defined. Violates property 1.Equation 3:The function 1/x is undefined at 0 but this function gives a value for g(0) namely -1 thus property 1 is not violated. The problem point is again the domain problem x = 0, so the question iswhat is theSince the limit is infinity, the limit doesn’t exist which violates property 2.
9 What type of functions are continuous at every point? A. Polynomial functionsB. Rational functions are continuous everywhere except where the denominator is 0.Theorem 4: The Intermediate Value TheoremIf f is a continuous function on a closed interval [a,b] and M is any number between f(a) and f(b) then there is at least one number c in [a,b] such that f(c ) = MTheorem 5: Existence of Zeros of a Continuous FunctionIf f is a continuous function on a closed interval [a,b] and if f(a) and f(b) have opposite signs then there is at least one solution of the equation f(x)=0 in the interval (a,b).
10 For Th. 4Pick a closed interval on the x axis say [1,2.5]f(1)=3 and f(2.5)=5So if I pick a y value between 3 and 5 (say 4) then there must be a value x between 1 and 2.5 such that f( c) = 4c = 2.25
11 For Th. 5Pick a closed interval on the x axis say[-2,0]f(-2)=-8 and f(0)=5Since -8 and 5 are opposite in sign the continuous graph must have crossed the x axis somewhere between -2 and 0. It looks like it could be at-1.5
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