Mathematics
1. Direct replacement in limit problems
We noted that when we solved limit problems limx?a f(x), sometimes
the final answer was just f(a).
Example.
limx?3
?
?x2 + x + 1 # $% & f(x)
?
? = limx?3
x2 + limx?3
x + limx?3
1
=
)
limx?3
x
*2
+3+1=32 +3+1 # $% & f(a)
.
We could have gotten the final answer by replacing x with a. In other
words the behavior near x = a, limx?a f(x) was determined by the value
f(a). We can actually do this for a large class of functions.
2
3
A function continuous at
x = 2, f(2) is defined, and
limx?2 f(x) exists, and
limx?2 f(x)=3= f(2).
2. Continuity – behavior near a number is determined by the
value at the number
2
3
A function not continuous
at x = 2, since f(2) is not
defined.
2
3
1
A function not continuous
at x = 2. Why? since
f(2) = 1 but
limx?2 f(x)=3.
Definition. A function f(x) is continuous at x = a if
In practice the graphs of functions that are continuous at x = a do not
have gaps or jumps at x = a. They usually can be sketched without lifting
off the pencil. However, there are exceptions (Problem 11).
2
3
An example of a function discontinuous (not continuous) at x = 2 since
limx?2 f(x) does not exist. (the left and right limits as x ? 2 are different:
limx?2- f(x) = 3 but limx?2+ f(x) = 1).
3. Removable discontinuities
If limx?a f(x) exists but limx?a f(x) ?= f(a), then f(x) has a removable
discontinuity at x = a. If we redefine f(a) to be
f(a) = limx?a f(x), then the new function is continuous at x = a. 1
2
For example,
f(x) = + x + 1 if x ?= 2
1.5 if x = 2
has a discontinuity at x = 2 since limx?2 f(x) = 3 but f(2) = 1.5. The
discontinuity x = 1.5 can be removed by redefining f(2) = 3.
4. Jump discontinuity
A jump discontinuity occurs at a number x = a if both limits limx?a- f(x)
and limx?a+ f(x) exist but they are different.
2
1.5
3
An example of a jump
discontinuity.
limx?2- f(x)=3 but
limx?2+ f(x)=1.5.
5. Classes of functions that are continuous
The first example in this section is a polynomial. In general,
Theorem 1 (Polynomials are continuous). Polynomials (i.e. f(x) = x3 –
x2 + 4 ) are functions continuous everywhere.
We prove this theorem using the definition of continuity and the Limits
Laws.
Let f(x) = c0 + c1x + ··· + cnxn. Then,
limx?a
f(x) = limx?a
(c0 + c1x + ··· + cnxn (1) )
= limx?a
c0 + limx?a
c1x + ··· + limx?a
cnxn (2)
= c0 + c1 limx?a
x + ··· + cn limx?a
xn (3)
= c0 + c1a + ··· + cnan (4) = f(a).
Without proof we state that also
Theorem 2. Rational functions (i.e f(x) = x3-x2+4
x2-x-4 ), exponential functions
(i.e. f(x) = ex), logarithmic functions (i.e. f(x) = ln x), root
functions ((i.e. f(x) = v3x) are all continous whenever they are defined.
Theorem 3. An inverse function of a continuous function is continuous.
Theorem 4. Also if f and g are continuous then f + g, f – g, f · g and
f/g are continuous functions whenever they are defined.
For example, let f(x) = x3 – x2 + 4 and g(x) = ex. Then both f and g
are continuous. Also x3 – x2 +4+ ex is continous, so are (x3 – x2 + 4)ex
and x3-x2+4
ex .
3
6. Composition
The limit of a continuous function can be computed by evaluating the
function at the limit. In other words the order of applying the limit and
the function can be changed.
Theorem 5 (A continuous function commutes with the limit). Let f be
continuous then,
limx?a
f(g(x)) = f(limx?a
g(x)).
This property is sometimes used when we solve limit problems. Also
note that if g is continuous then the right limit is just f(g(a)). So if we let
F = f(g(x)) then limx?a F(x) = F(a). We summarize this as
Theorem 6 (The composition of continuous functions is continuous). If
g is continuous at a number a and f is continuous at g(a), then F = f ? g
is continuous at a.
For example the function F(x) = v
x2 + 1 is continuous for all x since
it is the composition of f(x) = vx and g(x) = x2 + 1, both of which are
continuous on their domains.
We can also now detemine continuity of complicated functions built by
compositions and elementary operations without using the definition of the
limit. For example, we explain that the function f(x) = ln(x)+vx2 + 1·ex
is continuous in its domain x > 0.
Note that vx2 + 1 is continuous so is ex thus their product vx2 + 1 · ex
is continuous. Since ln(x) is continuous finally we conclude that f(x) is
continuous.
7. Intermediate Value Theorem
A key propetry of a continuous function f(x) on a = x = b is that it
assumes all values between f(a) and f(b). The continous function f(x)
cannot skip a number between f(a) and f(b) (otherwise there would be a
gap on the graph of f(x)).
a c b
f(a)
N
f(b)
Theorem 7 (Intermediate Value Theorem). Let f be continuous on the
interval [a, b]. Then for every value N between f(a) and f(b), there is some
c in [a, b] such that f(c) = N.
There are functions with a discontinuities that are neither removable nor
jump discontinuities. Can you sketch one?
4
8. Continuous from one side
a
f(a)
Continuous from the left. In the figure on the margin, f(x) is not continuous at x = a since
limx?a f(x) does not exist. However, as the values of x approach a from
the left, then the values of f(x) approach the value f(a). In this case,
Let f(x) be defined on an open interval I. If f is continuous at each
number in I then we say that f is continuous on the interval I.
For example let f(x) = v
1
x be defined on (0, 1). Since f is the fraction
of two functions, both of which are continuous on the interval (0, 1), and
since the denominator is not zero on (0, 1), f is a continuous function on
the interval (0, 1).
There is a subtlety what do if the domain of f is not open. For example,
let I = (0, 1]. Then we only evaluate limx?1- f(x) as f is not defined for
x > 1. Thus we can only determine whether limx?1- f(x) = f(1), that is
if f is continuous from the left at the right endpoint x = 1.
We agree then to say that a function f(x) defined on an interval I = (a, b]
is continuous on I if f is continuous at each number inside I and if f is
continuous from the right at x = b.
Similarly, we define continuity of intervals [a, b), [a, b], [a, 8), and (-8, b].