Continuity Theorems and Their use in Calculus

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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