# 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*are continuous on the interval (-infinity , +infinity).

^{ x}**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

\color{red}{\lim_{x\to\ a} g(x) = L}

*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*is continuous at

_{o}g*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. More on Questions on Continuity with Solutions

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