# 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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