Continuity & The I.V.T. & "Limits @ Infinity"
Continuity
Definition
: A function f is continuous at a number
a
if
. Otherwise, f is
discontinuous
at
a
or is said to have a
discontinuity
at
a
.
f is continuous
on the open interval I
if f is continuous at all
a
I.
Now, after analyzing a fancy sketch of a weird function on the blackboard you will see that f is continuous at a if and only if three conditions are satisfied (and these you memorize ):
1) f( a ) exists (i.e. a is in the domain of f)
2)
exists
3)
.
Definition
: f is continuous
from the right
at
a
(or
right continuous
at
a
) if
.
f is continuous from the left at
a
(or
left continuous
at
a
) if
.
f is continuous on the closed interval [c,d] if it is continuous on (c,d) and right continuous at c and left continuous at d.
Examples
1)
We have shown that
all polynomials
are continuous
everywhere
since we showed that if f is a polynomial then
.
2) Likewise, we have shown that all rational functions are continuous on their domains . (See the notes on limits.)
3) Discuss the continuity of the greates integer function
Recall: gif(x) = the largest integer less than or equal to x.
> gif:=floor(x);
> plot(gif,x=-3..3,discont=true,colour=red,thickness=2);
4) Discuss the continuity of the piecewise defined function
> pdf := piecewise(x < -1,x^2,-1 <= x and x<1 ,x+1,x>=1,-x^2+3);
> plot(pdf,x=-4..4,y=-4..4,discont=true,colour=red,thickness=2);
5)
Discuss the continuity of g(x) =
.
> g:=(x^3+1)/(x+1);
> plot(g,x=-4..4,colour=blue,thickness=2);
g(x) is continous everywhere except at x = -1, where it has a removeable discontinuity. Note: just because Maple's plot doesn't show a discontinuity it doesn't mean there isn't a discontinuity!!!
A function f has a
removeable discontinuity
at
a
if f is discontinuous at
a
but
exists. The discontinuity is "removed" by redefining f at
a
to be equal to its limit at
a
i.e. define f(
a
) =
.
e.g. Redefine the last example to be coninuous on R .
Intermediate Value Theorem (IVT)
IVT : (page 129 in Stewart ) If f is continuous on [a,b] and N is any number between f(a) and f(b) then there exists (at least one) a number c between a and b such that f(c) = N.
e.g. (#43, pg 132) Show that if
then there is a number
c
such that f(
c
) = 10.
>
> f:=x-> x^3-x^2+x;
> plot({f(x),10},x=-6..6,y=-15..15,colour=[red,blue],thickness=2);
Notice, from the picture it looks like c is about 2.4.
> f(2.4);
Not bad! Now, lets see what the IVT says. First we note that f(x) =
is
continuous
on all of R so the IVT applies to f on
every
interval we might be interested in.
> f(2);f(3);
From the above line we see that f(2) = 6 < 10 < f(3) = 21. Since f(2) < 10 < f(3) and f is continuous on [2,3] the IVT says that there exists a number c, 2 < c < 3, such that f(c) = 10 (as required by the problem!).
Note: the IVT doesn't tell us what the actual number is but only that it exists. Let's ask Maple to actually find c:
> solve(f(c)=10,c);
Wow!! (Be careful what you ask for!)
> evalf(%);
In other words, c is about 2.4 -- which we already knew!
e.g.'s
1) See e.g. #9 on page 130.
2)
Show that
has a
root
in (0,1).
3) Show that the cube root of three exists.
Limits @ Infinity -- Horizontal Asymptotes
Definition
: (pg. 134 in
Stewart
) If f is defined on (a,
) then
= L means that the value of f(x) can be made
arbitrarily close
to L by taking x sufficiently large. Simlar definition for
.
e.g.
By making a table of values or looking at a plot it is clear to see that
= 0 =
.
Definition
: If
= L
or
= M then the line y = L (or y = M) is a
horizontal asymptote
of the functon f.
Theorem
: If r > 0 then
and, if
is defined for all x < 0 then
.
To find limits @ infinity "by hand" it is often helpful to divide the numerator and denominator by a "convenient" power of the variable and then apply the limit laws together with the above theorem. Equivalent to this is factoring out the same power from the top and bottom and then cancelling. Try this on the examples below.
e.g.'s Find the HA's by Determining
and
.
> restart:with(plots):
1)
f(x) =
.
> f:=(2*x^4+x^3+1)/(1+x^2+3*x^4):
> Limit(f,x=infinity)=limit(f,x=infinity);Limit(f,x=-infinity)=limit(f,x=infinity);
> P1:=plot(f,x=-8..8,colour=red,thickness=2):P2:=plot(2/3,x=-8..8,colour=blue,thickness=2,linestyle=4):
> display(P1,P2);
2)
f(x) =
.
> f1:=sqrt(2*x^2+1)/(x+1):
> Limit(f1,x=-infinity)=limit(f1,x=-infinity);
> Limit(f1,x=infinity)=limit(f1,x=infinity);
> P1:=plot(f1,x=-6..6,y=-4..4,discont=true,colour=red,thickness=2):P2:=plot({sqrt(2),-sqrt(2)},x=-6..6,colour=blue,thickness=2,linestyle=4):
> display(P1,P2);
3)
Show that f(x) =
has no HA's.
> f2:=x^3-3*x^2+x-1;
> Limit(f2,x=-infinity)=limit(f2,x=-infinity);Limit(f2,x=infinity)=limit(f2,x=infinity);
> plot(f2,x=-2..4,y=-20..20,colour=red,thickness=2);