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 limits using continuity
Evaluate the following limits:
- \( \lim_{x\to 0} \sin (x) \)
- \( \lim_{x\to\pi} \cos (x) \)
- \( \lim_{x\to -1} \arctan(x) \)
Solution:
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: \( \lim_{x\to -1} \arctan(x) = \arctan(-1) = -\frac{\pi}{4} \)
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. \( (f \cdot g) \) is continuous at \( x = a \).
- D. \( (f / g) \) is continuous at \( x = a \) if \( g(a) \neq 0 \). (If \( g(a) = 0 \), then \( (f / g) \) is discontinuous at \( x = a \)).
Example: Continuity of algebraic combinations
Let \( f(x) = \sin x \) and \( g(x) = \cos x \). On which intervals are the functions \( (f + g) \), \( (f - g) \), \( (f \cdot g) \), and \( (f / g) \) continuous?
Solution:
Since both \( \sin x \) and \( \cos x \) are continuous everywhere, according to Theorem 2, \( (f + g) \), \( (f - g) \), and \( (f \cdot g) \) are continuous everywhere.
However, \( (f / g) \) is continuous everywhere except at values of \( x \) that make the denominator \( g(x) = 0 \). These values are found by solving:
\[ \cos x = 0 \]The values which make \( \cos x = 0 \) are:
\[ x = \frac{\pi}{2} + k\pi \quad \text{where } k \text{ is any integer.} \]Thus, \( (f / g) \) is continuous everywhere except at \( x = \frac{\pi}{2} + k\pi \).
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: Finding points of discontinuity
Find the values of \( x \) at which function \( f \) is discontinuous:
\[ f(x) = \frac{x - 2}{(2x^2 + 2x - 4)(x^4 + 5)} \]Solution:
The denominator of \( f \) is the product of two terms:
\[ (2x^2 + 2x - 4)(x^4 + 5) \]The term \( x^4 + 5 \) is always positive, hence never equal to zero. We now find the zeros of \( 2x^2 + 2x - 4 \) by solving:
\[ 2x^2 + 2x - 4 = 0 \]Dividing by 2 gives \( x^2 + x - 2 = 0 \), which factors into \( (x - 1)(x + 2) = 0 \).
The solutions are \( x = 1 \) and \( x = -2 \).
Therefore, function \( f \) is discontinuous at \( x = 1 \) and \( x = -2 \).
Theorem 4
If \( \lim_{x\to a} g(x) = L \) and if \( f \) is a continuous function at \( x = L \), then:
\[ \lim_{x\to a} f(g(x)) = f\left(\lim_{x\to a} g(x)\right) = f(L) \]Example: Evaluating composite limits
Evaluate the limit:
\[ \lim_{x\to a} \sin(2x + 5) \]Solution:
\( \sin x \) is continuous everywhere and \( 2x + 5 \) is a polynomial (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 \( f \circ g \) is continuous at \( x = a \).
Example: Proving continuity of exponential compositions
Show that any function of the form \( e^{ax + b} \) is continuous everywhere, where \( a \) and \( b \) are real numbers.
Solution:
The exponential function \( f(x) = e^x \) and the linear function \( g(x) = ax + b \) are continuous everywhere. Hence, their composition \( f(g(x)) = e^{ax + b} \) is also continuous everywhere.