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 \( (-\infty , +\infty) \).
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 \( g(x) = \cos x \). On which intervals 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:
\( x = \pi/2 + k \pi \) , where \( k \) is any integer.
\( (f / g) \) is continuous everywhere except at \( x = \pi/2 + k \pi \) , \( k \) is any 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 \( 2 x^2 + 2 x - 4 \) 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 \[ \lim_{x\to\ a} \sin(2x + 5) = \sin\left(\lim_{x\to\ a}(2x+5) \right) = \sin(2a + 5) \]


Theorem 5

If \( g \) is a continuous function at \( x = a \) and function \( f \) is continuous at \( g(a) \), then the composite function \( f_o g \) is continuous at \( x = a \).
Example:Show that any function of the form \( e^{ax + b} \) is continuous everywhere, \( a \) and \( b \) real numbers.
The exponential function\( f(x) = e^x \) and the polynomial (linear) function \( g(x) = a x + b \) are continuous everywhere. Hence the composition \( f(g(x)) = e^{ax + b} \) is also continuous everywhere.

More References and Links

Questions on Continuity with Solutions
Calculus Tutorials and Problems