If the problem says "graph", please graph. But you only
need to graph the function in the vicinity of the limiting $x$ value.
Don't forget to answer all parts of a problem.
I grade problems out of 2. Each homework is worth the
same, no matter how many I grade. So you are really getting a
percentage, not a total number of points.
I will drop (at least) two of your lowest homework and quiz grades.
Today we'll start section 1.8.
You have a new assignment -- some problems from the textbook,
which will be due next Monday. You also have some assignments
on IMath.
We ended with a dramatic example of the squeeze theorem. There are many
interesting strategies in math, and that theorem illustrates one: to
use two functions to demonstrate that a third has a limit.
What is the difference between continuous and discontinuous things?
A new administration in government?
A sidewalk?
Coffee poured into a cup?
The key is connectedness. Continuous things don't have to be
smooth, although we often think of them that way. But we certainly
don't want to encounter cliffs....
Definition: continuity at the point $x=a$:
\[
\lim_{x\rightarrow{a}}f(x)=f(a)
\]
Note: three things have to happen:
The limit of $f$ exists at $a$,
The function value $f(a)$ exists at $a$, and
The two values are equal.
Otherwise $f$ is discontinuous at $a$.
There are various kinds of discontinuity (which we've already seen):
Removable discontinuities (e.g. holes)
Jump discontinuities
Infinite discontinuities
"Crazy discontinuities" like the function of #59, p. 71:
$x^2$ for rational values of $x$, and 0 for
irrational ones.
Maybe a better definition:
A function is continuous on an interval if its graph can be traced there without lifting the pencil from the paper.
(Try one! Make one smooth, and one really "jaggy".) Only one thing: Make sure that you're drawing a function!
Many of the functions we've considered so far satisfy this, although not all do: consider the "holy function"
\[
f(x)=\frac{\cos(x)-1}{x^2}
\]
Open
Filled (and continuous): $f(0)\equiv -0.5$
This function has a limit at zero (-.5), but is not defined there. If
$f$ is not defined at $x=0$, then it cannot be continuous there. We can
fix this, by the way.... Just define $f(0)$ to be the limit (the
closed dot in the graph above).
Since continuity is wrapped up with limits, we have the same kinds
of results that we did in relation to limits:
We define one-sided continuity
we have sum laws, product, and quotient laws. We've
already used these to demonstrate that polynomials are
continuous on all real numbers:
\[
\lim_{x\to{a}}P(x)=P(a)
\]
Classes of functions that are continuous on their domains:
Polynomials
Rational functions: (ratios of polynomials)
Trigonometric functions: sine, cosine,
tangent
Root functions: $f(x)=x^{1/n}$
Exponential functions: $f(x)=b^x$
For all these functions one can evaluate limits using the
so-called "substitution method": simply plug in $a$ for
$x$.
An important theorem (p. 88), RE composite functions (remember
that I told you how important compositions are? Here's another place
they show up....):
Roughly: the composition of continuous functions is
continuous.
Let $F(x)=f(g(x))$ be a composite function.
If $g$ is continuous at $x=a$ and $f$ is continuous at $g(a)$
then
$F(x)$ is continuous at $x=a$:
\[
\lim\limits_{x\to a}F(x)=\lim\limits_{x\to a}f(g(x))=f(\lim\limits_{x\to a}g(x))=f(g(a))=F(a)
\]
That crazy function which was $x^2$ for rational values of $x$,
and 0 for irrational ones, has an interesting property: it is
continuous at only a single point! It satisfies the definition only at
$x=0$: it's discontinuous everywhere else! That's pretty weird.
The Intermediate Value Theorem is an important result of continuity:
Theorem: Suppose that f is continuous on the closed interval [a,b] and let N be any number between f(a) and f(b), where $f(a)\ne{f(b)}$. Then there exists a number $c$ in $(a,b)$ such that $f(c)=N$.
Example: pp. 93, #69
Website maintained by Andy Long.
Comments appreciated.
Updated on 01/22/2018 18:04:28
