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One-Sided Limits and Continuity
ByDr. Julia Arnold
One-Sided Limits
The function f has the right-hand limit L as x approaches a from the right written
L)x(flimax
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
M)x(flimax
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.
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
L)x(flimax L)x(flim)x(flim
axax 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
0ifx,x
0ifx,x)x(f
)x(flim0xFind Since = -0 =0 both the left and
right limits exist and are equal thus the limit is0.
0
Example 2: Let
0ifx,1
0ifx,1)x(g
)x(glim0xFind
1)x(glim0x The
1)x(glim0x
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.
Continuous Functions
A function f is continuous at the point x = a if the following conditions are satisfied.1. f(a) is defined.2. exists
3.
)x(flimax
)a(f)x(flimax
If a function is not continuous at a point then it is discontinuous.
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
A B
)(lim xfAx
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.
Equation 1:
1x1
1x2xxf
,
,)(
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 3xf1x )(lim
Hence, property 1 and 2 are okay.Property 3 is violated which requires that
13but
1fxf1x
...
)()(lim
F(x) is discontinuous at x=1because it violates the 3rd property
Equation 2: 2x4x
xf2
)( Since 2x2x
2x2x2x4x2
))((
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.
2
Equation 3:
0ifx1
0ifxx1
xg,
,)(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 the
)(lim xg0x
Since the limit is infinity, the limit doesn’t exist which violates property 2.
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 ) = M
Theorem 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).
x
yFor 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) = 4 c = 2.25
x
yFor 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
Now go to the homework.