Continuity
2.4
Most of the techniques of calculus require that functions
be continuous. A function is continuous if you can draw it
in one motion without picking up your pencil.
A function is continuous at a point if the limit is the same
as the value of the function.
This function has discontinuities
at x=1 and x=2.
2
1
1
2
3
4
It is continuous at x=0 and x=4,
because the one-sided limits
match the value of the function

Show g(x)=x^2 + 1 is continuous at x = 1
1) g (1)  2
2) lim g ( x)  2
x 1
3) lim g ( x)  g (1)  2
x 1
 g ( x) is continuous at x  1
 x 1 x  2
Is the function f(x)  
continuous at x  2?
2x - 1 x  2
1) f ( 2)  3
2) lim f ( x)  3
x 2
lim f ( x)  3
x 2
 lim f ( x) exists
x 2
3) lim f ( x)  f (2)  3
x2
 f ( x) is continuous at x  2
 x 1 x  2
Is the function f(x)  
continuous at x  2?
2x - 1 x  2
1) f (2)  DNE
 Not continuous at x  2
 x 1 x  2
 2
Is the function f(x)   x
x  2 continuous at x  2?
2 x  1 x  2

1) f (2)  4
2) lim f ( x)  3
x 2
lim f ( x)  3
x 2
 lim f ( x) exists
x 2
3) lim f ( x)  f (2)
x2
 f ( x) is discontinu ous at x  2
Types of Discontinuities

There are 4 types of discontinuities





Jump
Point
Essential
Removable
The first three are considered non
removable
Jump Discontinuity

Occurs when the curve breaks at a
particular point and starts somewhere else

Right hand limit does not equal left hand limit
Point Discontinuity

Occurs when the curve has a “hole”
because the function has a value that is
off the curve at that point.

Limit of f as x approaches x does not equal
f(x)
Essential Discontinuity

Occurs when curve has a vertical
asymptote

Limit dne due to asymptote
Removable Discontinuity

Occurs when you have a rational
expression with common factors in the
numerator and denominator. Because
these factors can be cancelled, the
discontinuity is removable.
Places to test for continuity

Rational Expression


Piecewise Functions


Changes in interval
Absolute Value Functions


Values that make denominator = 0
Use piecewise definition and test changes in
interval
Step Functions

Test jumps from 1 step to next.
Continuous Functions in their domains
Polynomials
 Rational f(x)/g(x) if g(x) ≠0
 Radical
 trig functions

Find and identify and points of
discontinuity
x  3 x  2
f(x)   2
x2
x
2) lim f ( x)  5
x 2
1) f ( 2)  5
lim f ( x)  4
x 2
 lim f ( x) dne
x 2
Non removable – jump discontinuity
Find and identify and points of
discontinuity
5
f ( x) 
x4
Non removable – essential discontinuity
VA at x = 4
Find and identify and points of
discontinuity
x  8 x  15
f ( x)  2
x  6x  5
2

x  5x  3
f ( x) 
x  5x  1
2 points of disc. (where denominator = 0)
Removable disc. At x = 5
Non removable essential at x = -1 (VA at x = -1)
Find and identify and points of
discontinuity
5
f(x)   2
x
x2
x2
1) f ( 2)  5
2) lim f ( x)  4
x2
lim f ( x)  4
x2
 lim f ( x)  4
x2
3) lim f ( x)  f (2)
x2
Non removable point discontinuity
Find and identify and points of
discontinuity
2 x  7 x  15
f ( x)  2
x  x  20
2

x  52 x  3
f ( x) 
x  5x  4
2 points of disc. (where denominator = 0)
Removable disc. At x = 5
Non removable essential at x = -4 (VA at x = -4)