Limits and Continuity
A graph that is continuous can be drawn without lifting your pencil.
The limit at any point will be defined at all points on a continuous graph
That means we are free to use our difference quotient to create an algebraic model for all secants
So far (intentionally), all of our graphs have been continuous…
Generally speaking, it does not really cause any problems for us…
The only thing we need to be careful of, however, is if P or Q are placed on the point of discontinuity….
Offence (km / h)
0 - 14 015 – 29 3 30 – 44 445 and over 6
Demerit (points)
Let x be the number of KM over
Let y be the number of demerit points
Offence (km / h)
0 < x < 15 y = 015 < x < 30 y = 3 30 < x < 45 y = 445 < x y =6
Demerit (points)
Create the following graph:
0 1 2 3 4
x
y
5
4
3
2
1
lim f(x) =
x 2-
1
lim f(x) =
x 2+
3
Therefore the function
is not continuous at x = 2
Algebraic limits
lim (5)
x -1
52 x
= 5
lim (3x4 – 5x)
x 2= 3(2)4 – 5(2)
= 38
lim
x -3=
=
5)3(2
1= dne
lim
x 12356
2
2
xxxx Always try to
factor first
lim
x 1 )1)(2()1)(5(
xxxx
)21()51(
=
=4
An indeterminate form:
lim
x 0 xx 24
Substitution will create division by zero, and we can not factor, so we will rationalize the numerator.
2424
xx
X
=41
)24(
44
xx
x
)24(
xx
x
24
1
x
Take the limit:
204
1
24
1
22
1
An indeterminate form: