Continuity Theorems and Their Applications in Calculus

Theorems, related to the continuity of functions and their applications in calculus are presented and discussed 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\to\0} \sin (x) \\\\ \lim_{x\to\pi} \cos (x) \\\\ \lim_{x\to\ -1} \arctan(x)

Solutions
If function f is continuous at x = a, then
\lim_{x\to\ a} \f(x) = f(a)

Since sin(x) is continuous
\lim_{x\to\0} \sin (x) = \sin(0) = 0

Since cos(x) is continuous
\lim_{x\to\pi} \cos (x) = \cos(\pi) = - 1

Since arctan(x) is continuous
\lim_{x\to\ -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 make 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 = π/2 + k(π) , where k is any integer.
(f / g) is continuous everywhere except at x = π/2 + k(π) , 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) = \dfrac{x-2}{(2 x^2 + 2 x - 4)(x^4 + 5)}

Solutions:
The denominator of f is the product of two terms and is given by
(2 x^2 + 2 x - 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 + 2 x - 4 = 0

The solutions are: x = 1 and x = - 2
function f is discontinuous at x = 1 and x = -2.

Theorem 4

If
\color{red}{\lim_{x\to\ a} g(x) = L}
and if f is a continuous function at x = L, then
\color{red}{\lim_{x\to\ a} f(g(x)) = f(\lim_{x\to\ a} g(x)) = f(L)}

Example:Evaluate the limit
\lim_{x\to\ a} \sin(2x + 5)

Solution: sin x is continuous everywhere and 2 x + 5 is a polynomial and also continuous everywhere. Hence
\lim_{x\to\ a} \sin(2x + 5) = \sin(\lim_{x\to\ a}(2x+5)) = \sin(2a + 5)

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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