We present an introduction and the definition of the concept of continuous functions in calculus with examples. Also, continuity theorems and their use in calculus are discussed.
Introduction and Definition of Continuous Functions
We first start with graphs of several continuous functions. The functions whose graphs are shown below are said to be continuous since these graphs have no "breaks", "gaps", or "holes".
We now present examples of discontinuous functions. These graphs have breaks, gaps, or points at which they are undefined.
Case 1: Undefined at a point (Hole)
In the graph below, the function is undefined at \( x = 2 \). The graph has a hole at \( x = 2 \) and the function is said to be discontinuous.
Case 2: Left and right limits do not match
In the graph below, the limits of the function to the left and to the right are not equal, and therefore the limit at \( x = 3 \) does not exist. The function is said to be discontinuous.
Case 3: Limit does not equal function value
The limit of the function at \( x = 2 \) exists, but it is not equal to the value of the function at \( x = 2 \). This function is also discontinuous.
Case 4: Vertical asymptote (Limit does not exist)
The limit of the function at \( x = 3 \) does not exist since to the left and to the right of 3 the function either increases or decreases indefinitely. This function is also discontinuous.
Taking into consideration all the information gathered from the examples of continuous and discontinuous functions shown above, we define a continuous function as follows:
Function \( f \) is continuous at a point \( a \) if the following three conditions are satisfied:
- 1. \( f(a) \) is defined
- 2. \( \lim_{x\to a} f(x) \) exists
- 3. \( \lim_{x\to a} f(x) = f(a) \)
Examples with Solutions
Example 1: Discontinuity due to division by zero
Show that function \( f \) defined below is not continuous at \( x = -2 \):
\[ f(x) = \frac{1}{x + 2} \]Solution:
\( f(-2) \) is undefined (division by 0 is not allowed); therefore, function \( f \) is discontinuous at \( x = -2 \).
Example 2: Continuity of a rational function
Show that function \( f \) is continuous for all values of \( x \) in \( \mathbb{R} \):
\[ f(x) = \frac{1}{x^4 + 6} \]Solution:
Function \( f \) is defined for all values of \( x \) in \( \mathbb{R} \). The limit of \( f \) at any point \( x = a \) is given by the quotient of two limits: the constant 1 and the limit of \( x^4 + 6 \), which is a polynomial function whose limit is \( a^4 + 6 \). Hence:
\[ \lim_{x\to a} f(x) = \frac{1}{a^4 + 6} \]Since \( f(a) = \frac{1}{a^4 + 6} \), we have:
\[ \lim_{x\to a} f(x) = f(a) \]The three conditions of continuity are satisfied, and therefore \( f \) is continuous for all values of \( x \) in \( \mathbb{R} \).
Example 3: Continuity of an absolute value function
Show that function \( f \) is continuous for all values of \( x \) in \( \mathbb{R} \):
\[ f(x) = |x - 5| \]Solution:
Let us write \( f(x) piecewise \):
- \( f(x) = x - 5 \) if \( x > 5 \)
- \( f(x) = -(x - 5) \) if \( x < 5 \)
- \( f(x) = 0 \) if \( x = 5 \)
\( f(x) \) is given by the polynomial functions \( x - 5 \) and \( -(x - 5) \) for \( x > 5 \) and \( x < 5 \) respectively, and hence \( f(x) \) is continuous for these values of \( x \).
We now check the point \( x = 5 \). Consider the limits as \( x \) approaches \( 5 \) from the left (\( x < 5 \)):
\[ \lim_{x\to 5^{-}} f(x) = \lim_{x\to 5^{-}} -(x - 5) = 0 \]Consider the limits as \( x \) approaches \( 5 \) from the right (\( x > 5 \)):
\[ \lim_{x\to 5^{+}} f(x) = \lim_{x\to 5^{+}} (x - 5) = 0 \]Since the one-sided limits are equal, the limit \( \lim_{x\to 5} f(x) \) exists and equals 0. Thus, \( \lim_{x\to 5} f(x) = 0 = f(5) \), and function \( f \) is continuous at \( x = 5 \). Combined with the intervals above, \( f \) is continuous for all values of \( x \) in \( \mathbb{R} \).