Continuity Theorems and Their use in Calculus

Theorems, related to the continuity of functions and their uses in calculus, are presented and dicussed with examples.

Theorem 1

All polynomial functions and the functions sin x, cos x, arctan x and e x are continuous on the interval (-infinity , +infinity).

Example:Evaluate the following limits:

lim
x→0 sin(x)
lim
x→ pi cos(x)
lim
x→ -1 arctan(x)

Solutions

Since sin(x) is continuous lim
x→ 0 sin(x) = sin (0) = 0
Since cos(x) is continuous lim
x→ pi cos(x) = cos (pi) = -1
arctan(x) is continuous lim
xx→ -1 arctan(x) = arctan (-1) = - pi / 4





Theorem 2

If functions f and g are continuous at x = a, then

A. (f + g) is continuous at x = a,

B. (f - g) is continuous at x = a,

C. (f . g) is continuous at x = a,

D. (f / g) is continuous at x = a if g(a) is not equal to zero.

If g(a) = 0 then (f / g) is discontinuous at x = a.


Example:Let f(x) = sin x and g(x) = cos x. Where are the following functions (f + g), (f - g), (f . g) and (f / g) continuous?

Solutions:

Since both sin x and cos x are continuous everywhere, according to theorem 2 above
(f + g), (f - g), (f . g) are continuous everywhere.

However
(f / g) is continuous everywhere except at values of x for which the denominator g(x) is equal to zero. These values are found by solving the trigonometric equation:
cos x = 0

The values which make
cos x = 0 are given by:
x = pi/2 + k(pi) , where k is any integer.

(f / g) is continuous everywhere except at x = pi/2 + k(pi) , k integer.

Theorem 3

A rational function is continuous everywhere except at the values of x that make the denominator of the function equal to zero.

Example:Find the values of x at which function f is discontinuous.

f(x) = (x - 2) / [ (2 x 2 + 2x - 4)(x 4 + 5) ]


Solutions:

The denominator of f is the product of two terms and is given by
(2 x 2 + 2x - 4)(x 4 + 5)

The term x
4 + 5 is always positive hence never equal to zero. We now need to find the zeros of 2 x 2 + 2x - 4 by solving the equation:
2 x
2 + 2x - 4 = 0

The solutions are: x = 1 and x = - 2

function f is discontinuous at x = 1 and x = -2.

Theorem 4

If limx→ ag(x) = L and if f is a continuous function at L, then limx→ af(g(x)) = f(limx→ ag(x)) = f(L).

Example:Evaluate the limit
limxx→ a sin (2x + 5)


Solution: sin x is continuous everywhere. Hence

lim
x a sin (2x + 5) = sin (limx a 2x + 5) = sin (2a + 5), 2x + 5 is a polynomial and therefore continuous everywhere.

Theorem 5

If g is a continuous function at x = a and function f is continuous at g(a), then the composition f o g is continuous at x = a.

Example:Show that any function of the form e ax + b is continuous everywhere, a and b real numbers.

f(x) = e
x the exponential function and g(x) = ax + b a polynomial (linear) function are continuous everywhere. Hence the composition f(g(x)) = e ax + b is also continuous everywhere.


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