Continuity

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Continuity

Definition
A function f is called continuous at a number a if

lim f (x) = f (a)

Otherwise, f is called discontinuous at a.

Note that the following must be satisfied:

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Geometrically, f is continuous at those points where its graph has no
breaks or holes

−1 0 1 2 3 4 5
Continuity x 3 / 16
On the previous figure, f is discontinuous at the points x = 1, x = 2,
x = 3, x = 4.
1 f is discontinuous at x = 1 since limx→1 f (x) does not exist.
2 f is discontinuous at x = 2 since f (2) is undefined.
3 f is discontinuous at x = 3 since limx→3 f (x) 6= f (3).
4 f is discontinuous at x = 4 since limx→4 f (x) does not exist and f (4)
is undefined.

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Type of discontinuities:
The discontinuity at x = 1 is called a jump discontinuity. (Here the
one-sided limits exist, but are not equal to each other).
The discontinuities at x = 2 and x = 3 are called removable
discontinuities, since f can be made continuous after we redefine it
appropriately at these points.
The discontinuity at x = 4 is called an infinite discontinuity.

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Definition
A function f is called continuous from the right at a if

lim f (x) = f (a).

Definition
A function f is called continuous from the left at a if

lim f (x) = f (a).

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Example

Heaviside function

0, if t < 0,
H(t) =
1, if t ≥ 0.
We have

lim H(t) = 1 = H(0),

so H is continuous from the right at t = 0.

lim H(t) = 0 6= H(0),

so H is discontinuous from the left at t = 0

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Definition
A function f is continuous on an interval if f is continuous at every
number of the interval.

At the endpoints of the interval “continuous” means ”continuous from the

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Continuity

Theorem
If f and g are continuous at a, and c is a constant, then the following
functions are also continuous at a:
1 c ·f
2 f + g,
3 f − g,
4 f · g,
5 f /g (if g (a) 6= 0).

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Limit of a composition

Theorem
Consider two functions, f and g , and a number a ∈ R.
Suppose that
the limit limx→a g (x) exists and is equal to a number b,
f is continuous at b.
Then
lim f (g (x)) = f (b).
x→a

In other words,  
lim f (g (x)) = f lim g (x) .
x→a x→a

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Theorem
The following functions are continuous at every number in their domains:

polynomials, rational functions, power functions, root functions,

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Continuity: Example
Problem
Where is the following function f continuous?
 2
x + 3x − ln x

f (x) = sin
tan−1 x − π/3

Solution
f is obtained from polynomial, trigonometric, inverse trigonometric, and
logarithmic functions using four basic arithmetic operations and
composition.
So, f is continuous everywhere on its domain.
We have to find the domain.
tan−1 and sin are defined everywhere.
ln x is defined when x > 0.
The denominator cannot be equal to 0: √
tan−1 x − π/3 √ 6
= 0 ⇔
√ tan−1 x 6= π/3 ⇔ x 6= tan(π/3) ⇔ x 6= 3
Answer: (0, 3) ∪ ( 3, ∞).
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Intermediate Value Theorem (IVT)

Theorem
Let f be a function continuous on a closed interval [a, b], such that
f (a) 6= f (b).
Suppose that N is a number strictly between f (a) and f (b), in other
words,
f (a) < N < f (b) or f (b) < N < f (a)
Then there exists a number c ∈ (a, b) such that f (c) = N.

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Intermediate Value Theorem (IVT)
y

f (b)

f (a)

a c b x

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Intermediate Value Theorem (IVT)
y

f (b)

f (a)

a c1 c2 c3 b x

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Intermediate value theorem: Example

Problem
Show that there is a root of the equation

5x 4 − 3x 3 + arctan(x 2 − 1) = 0

between 0 and 1.

Solution
f (x) = 5x 4 − 3x 3 + arctan(x 2 − 1) is continuous on [0, 1].
f (0) = arctan(−1) = − π4 < 0
f (1) = 5 − 3 + arctan(0) = 2 > 0
By the Intermediate Value Theorem with N = 0, there exists c ∈ (0, 1)
such that f (c) = 0, i.e., c is a root of the equation f (x) = 0.

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