1. One-Sided Limits and ContinuityBy
Dr. Julia Arnold

2. One-Sided Limits
The function f has the right-hand limit L as x approaches a from the right written
If 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 written
If 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. Then
If and only if
Thus the two-sided limit exists if and only if the one-sided limits exist and are equal.
Example 1: Let
Since = -0 =0 both the left and right limits exist and are equal thus the limit is 0.
Find

4. Example 2: Let
Find
The
Is the right hand limit.
Is the left hand limit
Since they are unequal the limit of g(x) as x approaches 0 does not exist.

5. Continuous Functions
A function f is continuous at the point x = a if the following conditions are satisfied.
1. f(a) is defined.
exists 3.
If a function is not continuous at a point then it is discontinuous.

6. A B
On 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 that
Equation 1:
F(x) is discontinuous at x=1
because it violates the 3rd property
Hence, property 1 and 2 are okay.
Property 3 is violated which requires that

8. Equation 2:
Since
We 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 is
what is the
Since 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 functions
B. Rational functions are continuous everywhere except where the denominator is 0.
Theorem 4: The Intermediate Value Theorem
If 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 ) = M
Theorem 5: Existence of Zeros of a Continuous Function
If 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. 4
Pick a closed interval on the x axis say [1,2.5]
f(1)=3 and f(2.5)=5
So 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) = 4
c = 2.25

11. For Th. 5
Pick a closed interval on the x axis say
[-2,0]
f(-2)=-8 and f(0)=5
Since -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

12. Now go to the homework.